problem solving involving quadratic equation

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Need Help Solving Those Dreaded Word Problems Involving Quadratic Equations?

Yes, I know it's tough. You've finally mastered factoring and using the quadratic formula and now you are asked to solve more problems!

Except these are even more tough. Now you have to figure out what the problem even means before trying to solve it. I completely understand and here's where I am going to try to help!

There are many types of problems that can easily be solved using your knowledge of quadratic equations. You may come across problems that deal with money and predicted incomes (financial) or problems that deal with physics such as projectiles. You may also come across construction type problems that deal with area or geometry problems that deal with right triangles.

Lucky for you, you can solve the quadratic equations, now you just have to learn how to apply this useful skill.

On this particular page, we are going to take a look at a physics "projectile problem".

Projectiles - Example 1

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t 2 + 40ft + 1.5. Find the maximum height attained by the ball.

Let's first take a minute to understand this problem and what it means. We know that a ball is being shot from a cannon. So, in your mind, imagine a cannon firing a ball. We know that the ball is going to shoot from the cannon, go into the air, and then fall to the ground.

So, here's a mathematical picture that I see in my head.

Now let's talk about what each part of this problem means. In our equation that we are given we must be given the value for the force of gravity (coefficient of t 2 ). We must also use our upward velocity (coefficient of t) and our original height of the cannon/ball (the constant or 1.5). Take a look...

Now that you have a mental picture of what's happening and you understand the formula given, we can go ahead and solve the problem.

First, ask yourself, "What am I solving for?" "What do I need to find?" You are asked to find the maximum height (go back and take a look at the diagram). What part of the parabola is this? Yes, it's the vertex! We will need to use the vertex formula and I will need to know the y coordinate of the vertex because it's asking for the height.
Next Step: Solve! Now that I know that I need to use the vertex formula, I can get to work.

Just as simple as that, this problem is solved.

Let's not stop here. Let's take this same problem and put a twist on it. There are many other things that we could find out about this ball!

Projectiles - Example 2

Same problem - different question. Take a look...

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t 2 + 40ft + 1.5. How long did it take for the ball to reach the ground?

Now, we've changed the question and we want to know how long did it take the ball to reach the ground.

What ground, you may ask. The problem didn't mention anything about a ground. Let's take a look at the picture "in our mind" again.

Do you see where the ball must fall to the ground. The x-axis is our "ground" in this problem. What do we know about points on the x-axis when we are dealing with quadratic equations and parabolas?

Yes, the points on the x-axis are our "zeros" or x-intercepts. This means that we must solve the quadratic equation in order to find the x-intercept.

Let's do it! Let's solve this equation. I'm thinking that this may not be a factorable equation. Do you agree? So, what's our solution?

Hopefully, you agree that we can use the quadratic formula to solve this equation.

The first time doesn't make sense because it's negative. This is the calculation for when the ball was on the ground initially before it was shot.

This actually never really occurred because the ball was shot from the cannon and was never shot from the ground. Therefore, we will disregard this answer.

The other answer was 2.54 seconds which is when the ball reached the ground (x-axis) after it was shot. Therefore, this is the only correct answer to this problem.

Ok, one more spin on this problem. What would you do in this case?

Projectiles - Example 3

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t 2 + 40ft + 1.5. How long does it take the ball to reach a height of 20 feet?

Yes, this problem is a little trickier because the question is not asking for the maximum height (vertex) or the time it takes to reach the ground (zeros), instead it it asking for the time it takes to reach a height of 20 feet.

Since the ball reaches a maximum height of 26.5 ft, we know that it will reach a height of 20 feet on the way up and on the way down.

Let's just estimate on our graph and also make sure that we get this visual in our head.

From looking at this graph, I would estimate the times to be about 0.7 sec and 1.9 sec. Do you see how the ball will reach 20 feet on the way up and on the way down?

Now, let's find the actual values. Where will we substitute 20 feet?

Yes, we must substitute 20 feet for h(t) because this is the given height. We will now be solving for t using the quadratic formula. Take a look.

Our actual times were pretty close to our estimates. Just don't forget that when you solve a quadratic equation, you must have the equation set equal to 0. Therefore, we had to subtract 20 from both sides in order to have the equation set to 0.

You've now seen it all when it comes to projectiles!

Great Job! Hopefully you've been able to understand how to solve problems involving quadratic equations. I also hope that you better understand these common velocity equations and how to think about what this problem looks like graphically in order to help you to understand which process or formula to use in order to solve the problem.

Quadratic Equations
Projectile Problems

problem solving involving quadratic equation

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9.6: Solve Applications of Quadratic Equations

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Learning Objectives

By the end of this section, you will be able to:

Solve applications modeled by quadratic equations

Be Prepared 9.13

Before you get started, take this readiness quiz.

The sum of two consecutive odd numbers is −100. Find the numbers. If you missed this problem, review Example 2.18.

Be Prepared 9.14

Solve: 2 x + 1 + 1 x − 1 = 1 x 2 − 1 . 2 x + 1 + 1 x − 1 = 1 x 2 − 1 . If you missed this problem, review Example 7.35.

Be Prepared 9.15

Find the length of the hypotenuse of a right triangle with legs 5 inches and 12 inches. If you missed this problem, review Example 2.34.

Solve Applications Modeled by Quadratic Equations

We solved some applications that are modeled by quadratic equations earlier, when the only method we had to solve them was factoring. Now that we have more methods to solve quadratic equations, we will take another look at applications.

Let’s first summarize the methods we now have to solve quadratic equations.

Methods to Solve Quadratic Equations

Square Root Property
Completing the Square
Quadratic Formula

As you solve each equation, choose the method that is most convenient for you to work the problem. As a reminder, we will copy our usual Problem-Solving Strategy here so we can follow the steps.

Use a Problem-Solving Strategy.

Step 1. Read the problem. Make sure all the words and ideas are understood.
Step 2. Identify what we are looking for.
Step 3. Name what we are looking for. Choose a variable to represent that quantity.
Step 4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
Step 5. Solve the equation using algebra techniques.
Step 6. Check the answer in the problem and make sure it makes sense.
Step 7. Answer the question with a complete sentence

We have solved number applications that involved consecutive even and odd integers, by modeling the situation with linear equations. Remember, we noticed each even integer is 2 more than the number preceding it. If we call the first one n , then the next one is n + 2. The next one would be n + 2 + 2 or n + 4. This is also true when we use odd integers. One set of even integers and one set of odd integers are shown below.

Consecutive even integers Consecutive odd integers 64 , 66 , 68 77 , 79 , 81 n 1 st even integer n 1 st odd integer n + 2 2 nd consecutive even integer n + 2 2 nd consecutive odd integer n + 4 3 rd consecutive even integer n + 4 3 rd consecutive odd integer Consecutive even integers Consecutive odd integers 64 , 66 , 68 77 , 79 , 81 n 1 st even integer n 1 st odd integer n + 2 2 nd consecutive even integer n + 2 2 nd consecutive odd integer n + 4 3 rd consecutive even integer n + 4 3 rd consecutive odd integer

Some applications of odd or even consecutive integers are modeled by quadratic equations. The notation above will be helpful as you name the variables.

Example 9.35

The product of two consecutive odd integers is 195. Find the integers.

Try It 9.69

The product of two consecutive odd integers is 99. Find the integers.

Try It 9.70

The product of two consecutive even integers is 168. Find the integers.

We will use the formula for the area of a triangle to solve the next example.

Area of a Triangle

For a triangle with base, b , and height, h , the area, A , is given by the formula A = 1 2 b h . A = 1 2 b h .

$Image of a trangle. The horizontal base side is labeled b, and a line segment labeled h is perpendicular to the base, connecting it to the opposite vertex.$

Recall that when we solve geometric applications, it is helpful to draw the figure.

Example 9.36

An architect is designing the entryway of a restaurant. She wants to put a triangular window above the doorway. Due to energy restrictions, the window can only have an area of 120 square feet and the architect wants the base to be 4 feet more than twice the height. Find the base and height of the window.

Try It 9.71

Find the base and height of a triangle whose base is four inches more than six times its height and has an area of 456 square inches.

Try It 9.72

If a triangle that has an area of 110 square feet has a base that is two feet less than twice the height, what is the length of its base and height?

In the two preceding examples, the number in the radical in the Quadratic Formula was a perfect square and so the solutions were rational numbers. If we get an irrational number as a solution to an application problem, we will use a calculator to get an approximate value.

We will use the formula for the area of a rectangle to solve the next example.

Area of a Rectangle

For a rectangle with length, L , and width, W , the area, A , is given by the formula A = LW .

$Image shows a rectangle. All four angles are marked as right angles. The longer, horizontal side is labeled L and the shorter, vertical side is labeled w.$

Example 9.37

Mike wants to put 150 square feet of artificial turf in his front yard. This is the maximum area of artificial turf allowed by his homeowners association. He wants to have a rectangular area of turf with length one foot less than 3 times the width. Find the length and width. Round to the nearest tenth of a foot.

Try It 9.73

The length of a 200 square foot rectangular vegetable garden is four feet less than twice the width. Find the length and width of the garden, to the nearest tenth of a foot.

Try It 9.74

A rectangular tablecloth has an area of 80 square feet. The width is 5 feet shorter than the length.What are the length and width of the tablecloth to the nearest tenth of a foot.?

The Pythagorean Theorem gives the relation between the legs and hypotenuse of a right triangle. We will use the Pythagorean Theorem to solve the next example.

Pythagorean Theorem

In any right triangle, where a and b are the lengths of the legs, and c is the length of the hypotenuse, a 2 + b 2 = c 2 .

$Image shows a right triangle with horizontal and vertical legs. The vertical leg is labeled a. The horizontal side is labeled b. The hypotenuse is labeled c.$

Example 9.38

Rene is setting up a holiday light display. He wants to make a ‘tree’ in the shape of two right triangles, as shown below, and has two 10-foot strings of lights to use for the sides. He will attach the lights to the top of a pole and to two stakes on the ground. He wants the height of the pole to be the same as the distance from the base of the pole to each stake. How tall should the pole be?

Try It 9.75

The sun casts a shadow from a flag pole. The height of the flag pole is three times the length of its shadow. The distance between the end of the shadow and the top of the flag pole is 20 feet. Find the length of the shadow and the length of the flag pole. Round to the nearest tenth.

Try It 9.76

The distance between opposite corners of a rectangular field is four more than the width of the field. The length of the field is twice its width. Find the distance between the opposite corners. Round to the nearest tenth.

The height of a projectile shot upward from the ground is modeled by a quadratic equation. The initial velocity, v 0 , propels the object up until gravity causes the object to fall back down.

Projectile motion

The height in feet, h , of an object shot upwards into the air with initial velocity, v 0 v 0 , after t t seconds is given by the formula

h = −16 t 2 + v 0 t h = −16 t 2 + v 0 t

We can use this formula to find how many seconds it will take for a firework to reach a specific height.

Example 9.39

A firework is shot upwards with initial velocity 130 feet per second. How many seconds will it take to reach a height of 260 feet? Round to the nearest tenth of a second.

Try It 9.77

An arrow is shot from the ground into the air at an initial speed of 108 ft/s. Use the formula h = −16 t 2 + v 0 t to determine when the arrow will be 180 feet from the ground. Round the nearest tenth.

Try It 9.78

A man throws a ball into the air with a velocity of 96 ft/s. Use the formula h = −16 t 2 + v 0 t to determine when the height of the ball will be 48 feet. Round to the nearest tenth.

We have solved uniform motion problems using the formula D = rt in previous chapters. We used a table like the one below to organize the information and lead us to the equation.

$Image shows the template for a table with three rows and four columns. The first column is empty. The second column is labeled “Rate.” The third column is labeled “Time.” The fourth column is labeled “Distance.” The labels are written in the equation Rate times Time equals Distance. There is one extra cell at the bottom of the fourth column.$

The formula D = rt assumes we know r and t and use them to find D . If we know D and r and need to find t , we would solve the equation for t and get the formula t = D r . t = D r .

Some uniform motion problems are also modeled by quadratic equations.

Example 9.40

Professor Smith just returned from a conference that was 2,000 miles east of his home. His total time in the airplane for the round trip was 9 hours. If the plane was flying at a rate of 450 miles per hour, what was the speed of the jet stream?

This is a uniform motion situation. A diagram will help us visualize the situation.

Try It 9.79

MaryAnne just returned from a visit with her grandchildren back east . The trip was 2400 miles from her home and her total time in the airplane for the round trip was 10 hours. If the plane was flying at a rate of 500 miles per hour, what was the speed of the jet stream?

Try It 9.80

Gerry just returned from a cross country trip. The trip was 3000 miles from his home and his total time in the airplane for the round trip was 11 hours. If the plane was flying at a rate of 550 miles per hour, what was the speed of the jet stream?

Work applications can also be modeled by quadratic equations. We will set them up using the same methods we used when we solved them with rational equations.We’ll use a similar scenario now.

Example 9.41

The weekly gossip magazine has a big story about the presidential election and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes 12 hours more than Press #2 to do the job and when both presses are running they can print the job in 8 hours. How long does it take for each press to print the job alone?

This is a work problem. A chart will help us organize the information.

We are looking for how many hours it would take each press separately to complete the job.

Try It 9.81

The weekly news magazine has a big story naming the Person of the Year and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes 6 hours more than Press #2 to do the job and when both presses are running they can print the job in 4 hours. How long does it take for each press to print the job alone?

Try It 9.82

Erlinda is having a party and wants to fill her hot tub. If she only uses the red hose it takes 3 hours more than if she only uses the green hose. If she uses both hoses together, the hot tub fills in 2 hours. How long does it take for each hose to fill the hot tub?

Access these online resources for additional instruction and practice with solving applications modeled by quadratic equations.

Word Problems Involving Quadratic Equations
Quadratic Equation Word Problems
Applying the Quadratic Formula

Section 9.5 Exercises

Practice makes pefect.

In the following exercises, solve using any method.

The product of two consecutive odd numbers is 255. Find the numbers.

The product of two consecutive even numbers is 360. Find the numbers.

The product of two consecutive even numbers is 624. Find the numbers.

The product of two consecutive odd numbers is 1,023. Find the numbers.

The product of two consecutive odd numbers is 483. Find the numbers.

The product of two consecutive even numbers is 528. Find the numbers.

In the following exercises, solve using any method. Round your answers to the nearest tenth, if needed.

A triangle with area 45 square inches has a height that is two less than four times the base Find the base and height of the triangle.

The base of a triangle is six more than twice the height. The area of the triangle is 88 square yards. Find the base and height of the triangle.

The area of a triangular flower bed in the park has an area of 120 square feet. The base is 4 feet longer that twice the height. What are the base and height of the triangle?

A triangular banner for the basketball championship hangs in the gym. It has an area of 75 square feet. What is the length of the base and height , if the base is two-thirds of the height?

The length of a rectangular driveway is five feet more than three times the width. The area is 50 square feet. Find the length and width of the driveway.

A rectangular lawn has area 140 square yards. Its width that is six less than twice the length. What are the length and width of the lawn?

A rectangular table for the dining room has a surface area of 24 square feet. The length is two more feet than twice the width of the table. Find the length and width of the table.

The new computer has a surface area of 168 square inches. If the the width is 5.5 inches less that the length, what are the dimensions of the computer?

The hypotenuse of a right triangle is twice the length of one of its legs. The length of the other leg is three feet. Find the lengths of the three sides of the triangle.

The hypotenuse of a right triangle is 10 cm long. One of the triangle’s legs is three times the length of the other leg. Find the lengths of the two legs of the triangle. Round to the nearest tenth.

A rectangular garden will be divided into two plots by fencing it on the diagonal. The diagonal distance from one corner of the garden to the opposite corner is five yards longer than the width of the garden. The length of the garden is three times the width. Find the length of the diagonal of the garden.

$Image shows a rectangular segment of grass with fence around 4 sides and across the diagonal. The vertical side of the rectangle is labeled w and the horizontal side is labeled 3 w. The diagonal fence is labeled w plus 5.$

Nautical flags are used to represent letters of the alphabet. The flag for the letter, O consists of a yellow right triangle and a red right triangle which are sewn together along their hypotenuse to form a square. The hypotenuse of the two triangles is three inches longer than a side of the flag. Find the length of the side of the flag.

$Image shows a square with side lengths s. The square is divided into two triangles with a diagonal. The top triangle is red and the lower triangle is yellow. The diagonal is labeled s plus 3.$

Gerry plans to place a 25-foot ladder against the side of his house to clean his gutters. The bottom of the ladder will be 5 feet from the house.How far up the side of the house will the ladder reach?

John has a 10-foot piece of rope that he wants to use to support his 8-foot tree. How far from the base of the tree should he secure the rope?

A firework rocket is shot upward at a rate of 640 ft/sec. Use the projectile formula h = −16 t 2 + v 0 t to determine when the height of the firework rocket will be 1200 feet.

An arrow is shot vertically upward at a rate of 220 feet per second. Use the projectile formula h = −16 t 2 + v 0 t , to determine when height of the arrow will be 400 feet.

A bullet is fired straight up from a BB gun with initial velocity 1120 feet per second at an initial height of 8 feet. Use the formula h = −16 t 2 + v 0 t + 8 to determine how many seconds it will take for the bullet to hit the ground. (That is, when will h = 0?)

A stone is dropped from a 196-foot platform. Use the formula h = −16 t 2 + v 0 t + 196 to determine how many seconds it will take for the stone to hit the ground. (Since the stone is dropped, v 0 = 0.)

The businessman took a small airplane for a quick flight up the coast for a lunch meeting and then returned home. The plane flew a total of 4 hours and each way the trip was 200 miles. What was the speed of the wind that affected the plane which was flying at a speed of 120 mph?

The couple took a small airplane for a quick flight up to the wine country for a romantic dinner and then returned home. The plane flew a total of 5 hours and each way the trip was 300 miles. If the plane was flying at 125 mph, what was the speed of the wind that affected the plane?

Roy kayaked up the river and then back in a total time of 6 hours. The trip was 4 miles each way and the current was difficult. If Roy kayaked at a speed of 5 mph, what was the speed of the current?

Rick paddled up the river, spent the night camping, and and then paddled back. He spent 10 hours paddling and the campground was 24 miles away. If Rick kayaked at a speed of 5 mph, what was the speed of the current?

Two painters can paint a room in 2 hours if they work together. The less experienced painter takes 3 hours more than the more experienced painter to finish the job. How long does it take for each painter to paint the room individually?

Two gardeners can do the weekly yard maintenance in 8 minutes if they work together. The older gardener takes 12 minutes more than the younger gardener to finish the job by himself. How long does it take for each gardener to do the weekly yard maintainence individually?

It takes two hours for two machines to manufacture 10,000 parts. If Machine #1 can do the job alone in one hour less than Machine #2 can do the job, how long does it take for each machine to manufacture 10,000 parts alone?

Sully is having a party and wants to fill his swimming pool. If he only uses his hose it takes 2 hours more than if he only uses his neighbor’s hose. If he uses both hoses together, the pool fills in 4 hours. How long does it take for each hose to fill the pool?

Writing Exercises

Make up a problem involving the product of two consecutive odd integers.

ⓐ Start by choosing two consecutive odd integers. What are your integers?

ⓑ What is the product of your integers?

ⓓ Did you get the numbers you started with?

Make up a problem involving the product of two consecutive even integers.

ⓐ Start by choosing two consecutive even integers. What are your integers?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

$This table provides a checklist to evaluate mastery of the objectives of this section. Choose how would you respond to the statement “I can solve applications of the quadratic formula.” “Confidently,” “with some help,” or “No, I don’t get it.”$

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Real World Examples of Quadratic Equations

A Quadratic Equation looks like this:

Quadratic equations pop up in many real world situations!

Here we have collected some examples for you, and solve each using different methods:

Factoring Quadratics
Completing the Square
Graphing Quadratic Equations
The Quadratic Formula
Online Quadratic Equation Solver

Each example follows three general stages:

Take the real world description and make some equations
Use your common sense to interpret the results

Balls, Arrows, Missiles and Stones

When you throw a ball (or shoot an arrow, fire a missile or throw a stone) it goes up into the air, slowing as it travels, then comes down again faster and faster ...

... and a Quadratic Equation tells you its position at all times!

Example: Throwing a Ball

A ball is thrown straight up, from 3 m above the ground, with a velocity of 14 m/s. when does it hit the ground.

Ignoring air resistance, we can work out its height by adding up these three things: (Note: t is time in seconds)

Add them up and the height h at any time t is:

h = 3 + 14t − 5t 2

And the ball will hit the ground when the height is zero:

3 + 14t − 5t 2 = 0

Which is a Quadratic Equation !

In "Standard Form" it looks like:

−5t 2 + 14t + 3 = 0

It looks even better when we multiply all terms by −1 :

5t 2 − 14t − 3 = 0

Let us solve it ...

There are many ways to solve it, here we will factor it using the "Find two numbers that multiply to give ac , and add to give b " method in Factoring Quadratics :

ac = −15 , and b = −14 .

The factors of −15 are: −15, −5, −3, −1, 1, 3, 5, 15

By trying a few combinations we find that −15 and 1 work (−15×1 = −15, and −15+1 = −14)

The "t = −0.2" is a negative time, impossible in our case.

The "t = 3" is the answer we want:

The ball hits the ground after 3 seconds!

Here is the graph of the Parabola h = −5t 2 + 14t + 3

It shows you the height of the ball vs time

Some interesting points:

(0,3) When t=0 (at the start) the ball is at 3 m

(−0.2,0) says that −0.2 seconds BEFORE we threw the ball it was at ground level. This never happened! So our common sense says to ignore it.

(3,0) says that at 3 seconds the ball is at ground level.

Also notice that the ball goes nearly 13 meters high.

Note: You can find exactly where the top point is!

The method is explained in Graphing Quadratic Equations , and has two steps:

Find where (along the horizontal axis) the top occurs using −b/2a :

t = −b/2a = −(−14)/(2 × 5) = 14/10 = 1.4 seconds

Then find the height using that value (1.4)

h = −5t 2 + 14t + 3 = −5(1.4) 2 + 14 × 1.4 + 3 = 12.8 meters

So the ball reaches the highest point of 12.8 meters after 1.4 seconds.

Example: New Sports Bike

You have designed a new style of sports bicycle!

Now you want to make lots of them and sell them for profit.

Your costs are going to be:

$700,000 for manufacturing set-up costs, advertising, etc
$110 to make each bike

Based on similar bikes, you can expect sales to follow this "Demand Curve":

Where "P" is the price.

For example, if you set the price:

at $0, you just give away 70,000 bikes
at $350, you won't sell any bikes at all
at $300 you might sell 70,000 − 200×300 = 10,000 bikes

So ... what is the best price? And how many should you make?

Let us make some equations!

How many you sell depends on price, so use "P" for Price as the variable

Profit = −200P 2 + 92,000P − 8,400,000

Yes, a Quadratic Equation. Let us solve this one by Completing the Square .

Solve: −200P 2 + 92,000P − 8,400,000 = 0

Step 1 Divide all terms by -200

Step 2 Move the number term to the right side of the equation:

Step 3 Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation:

(b/2) 2 = (−460/2) 2 = (−230) 2 = 52900

Step 4 Take the square root on both sides of the equation:

Step 5 Subtract (-230) from both sides (in other words, add 230):

What does that tell us? It says that the profit is ZERO when the Price is $126 or $334

But we want to know the maximum profit, don't we?

It is exactly half way in-between! At $230

And here is the graph:

The best sale price is $230 , and you can expect:

Unit Sales = 70,000 − 200 x 230 = 24,000
Sales in Dollars = $230 x 24,000 = $5,520,000
Costs = 700,000 + $110 x 24,000 = $3,340,000
Profit = $5,520,000 − $3,340,000 = $2,180,000

A very profitable venture.

Example: Small Steel Frame

Your company is going to make frames as part of a new product they are launching.

The frame will be cut out of a piece of steel, and to keep the weight down, the final area should be 28 cm 2

The inside of the frame has to be 11 cm by 6 cm

What should the width x of the metal be?

Area of steel before cutting:

Area of steel after cutting out the 11 × 6 middle:

Let us solve this one graphically !

Here is the graph of 4x 2 + 34x :

The desired area of 28 is shown as a horizontal line.

The area equals 28 cm 2 when:

x is about −9.3 or 0.8

The negative value of x make no sense, so the answer is:

x = 0.8 cm (approx.)

Example: River Cruise

A 3 hour river cruise goes 15 km upstream and then back again. the river has a current of 2 km an hour. what is the boat's speed and how long was the upstream journey.

There are two speeds to think about: the speed the boat makes in the water, and the speed relative to the land:

Let x = the boat's speed in the water (km/h)
Let v = the speed relative to the land (km/h)

Because the river flows downstream at 2 km/h:

when going upstream, v = x−2 (its speed is reduced by 2 km/h)
when going downstream, v = x+2 (its speed is increased by 2 km/h)

We can turn those speeds into times using:

time = distance / speed

(to travel 8 km at 4 km/h takes 8/4 = 2 hours, right?)

And we know the total time is 3 hours:

total time = time upstream + time downstream = 3 hours

Put all that together:

total time = 15/(x−2) + 15/(x+2) = 3 hours

Now we use our algebra skills to solve for "x".

First, get rid of the fractions by multiplying through by (x-2) (x+2) :

3(x-2)(x+2) = 15(x+2) + 15(x-2)

Expand everything:

3(x 2 −4) = 15x+30 + 15x−30

Bring everything to the left and simplify:

3x 2 − 30x − 12 = 0

It is a Quadratic Equation!

Let us solve it using the Quadratic Formula :

Where a , b and c are from the Quadratic Equation in "Standard Form": ax 2 + bx + c = 0

Solve 3x 2 - 30x - 12 = 0

Answer: x = −0.39 or 10.39 (to 2 decimal places)

x = −0.39 makes no sense for this real world question, but x = 10.39 is just perfect!

Answer: Boat's Speed = 10.39 km/h (to 2 decimal places)

And so the upstream journey = 15 / (10.39−2) = 1.79 hours = 1 hour 47min

And the downstream journey = 15 / (10.39+2) = 1.21 hours = 1 hour 13min

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A quadratic equation has two solutions if the discriminant b^2 - 4ac is positive.

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Module 10: Quadratic Equations and Functions

10.3 – applications of quadratic functions, learning objectives, objects in free fall.

Determining the width of a border

Finding the maximum and minimum values of a quadratic function

(10.3.1) – solve application problems involving quadratic functions.

Quadratic equations are widely used in science, business, and engineering. Quadratic equations are commonly used in situations where two things are multiplied together and they both depend on the same variable. For example, when working with area, if both dimensions are written in terms of the same variable, you use a quadratic equation. Because the quantity of a product sold often depends on the price, you sometimes use a quadratic equation to represent revenue as a product of the price and the quantity sold. Quadratic equations are also used when gravity is involved, such as the path of a ball or the shape of cables in a suspension bridge.

A very common and easy-to-understand application is the height of a ball thrown at the ground off a building. Because gravity will make the ball speed up as it falls, a quadratic equation can be used to estimate its height any time before it hits the ground. Note: The equation isn’t completely accurate, because friction from the air will slow the ball down a little. For our purposes, this is close enough.

A ball is thrown off a building from 200 feet above the ground. Its starting velocity (also called initial velocity ) is [latex]−10[/latex] feet per second. (The negative value means it’s heading toward the ground.)

The equation [latex]h=-16t^{2}-10t+200[/latex] can be used to model the height of the ball after [latex]t[/latex] seconds. About how long does it take for the ball to hit the ground?

When the ball hits the ground, the height is 0. Substitute 0 for [latex]h[/latex].

[latex]\begin{array}{c}h=-16t^{2}-10t+200\\0=-16t^{2}-10t+200\\-16t^{2}-10t+200=0\end{array}[/latex]

This equation is difficult to solve by factoring or by completing the square, so solve it by applying the Quadratic Formula, [latex] x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}[/latex]. In this case, the variable is [latex]t[/latex] rather than [latex]x[/latex]. [latex]a=−16,b=−10[/latex], and [latex]c=200[/latex].

[latex]\displaystyle t=\frac{-(-10)\pm \sqrt{{{(-10)}^{2}}-4(-16)(200)}}{2(-16)}[/latex]

Simplify. Be very careful with the signs.

[latex]\large \begin{array}{l}t=\frac{10\pm \sqrt{100+12800}}{-32}\\\,\,=\frac{10\pm \sqrt{12900}}{-32}\end{array}[/latex]

Use a calculator to find both roots.

[latex]t[/latex] is approximately [latex]−3.86[/latex] or [latex]3.24[/latex].

Consider the roots logically. One solution, [latex]−3.86[/latex], cannot be the time because it is a negative number. The other solution, [latex]3.24[/latex] seconds, must be when the ball hits the ground.

The ball hits the ground approximately [latex]3.24[/latex] seconds after being thrown.

In the next video we show another example of how the quadratic equation can be used to find the time it takes for an object in free fall to hit the ground.

Here are some more similar objects in free fall examples.

Example: Applying the Vertex and [latex] x[/latex] -Intercepts of a Parabola

A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation [latex]H\left(t\right)=-16{t}^{2}+80t+40[/latex].

a. When does the ball reach the maximum height?

b. What is the maximum height of the ball?

c. When does the ball hit the ground?

[latex]\large \begin{array}{c} h=-\frac{80}{2\left(-16\right)} \text{ }=\frac{80}{32}\hfill \\ \text{ }=\frac{5}{2}\hfill \\ \text{ }=2.5\hfill \end{array}[/latex]

The ball reaches a maximum height after 2.5 seconds.

b. To find the maximum height, find the y coordinate of the vertex of the parabola.

[latex]\large \begin{array}{c}k=H\left(-\frac{b}{2a}\right)\hfill \\ \text{ }=H\left(2.5\right)\hfill \\ \text{ }=-16{\left(2.5\right)}^{2}+80\left(2.5\right)+40\hfill \\ \text{ }=140\hfill \end{array}[/latex]

The ball reaches a maximum height of 140 feet.

c. To find when the ball hits the ground, we need to determine when the height is zero, [latex]H\left(t\right)=0[/latex].

We use the quadratic formula.

[latex]\large \begin{array}{c} t=\frac{-80\pm \sqrt{{80}^{2}-4\left(-16\right)\left(40\right)}}{2\left(-16\right)}\hfill \\ \text{ }=\frac{-80\pm \sqrt{8960}}{-32}\hfill \end{array}[/latex]

Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.

[latex]\large \begin{array}{c}t=\frac{-80-\sqrt{8960}}{-32}\approx 5.458\hfill & \text{or}\hfill & t=\frac{-80+\sqrt{8960}}{-32}\approx -0.458\hfill \end{array}[/latex]

The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds.

Graph of a negative parabola where x goes from -1 to 6.

A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock’s height above ocean can be modeled by the equation [latex]H\left(t\right)=-16{t}^{2}+96t+112[/latex].

a. When does the rock reach the maximum height?

b. What is the maximum height of the rock?

c. When does the rock hit the ocean?

a. 3 seconds

b. 256 feet

c. 7 seconds

Applications of quadratic functions: determining the width of a border

The area problem below does not look like it includes a Quadratic Formula of any type, and the problem seems to be something you have solved many times before by simply multiplying. But in order to solve it, you will need to use a quadratic equation.

Bob made a quilt that is 4 ft [latex]\times[/latex] 5 ft. He has 10 sq. ft. of fabric he can use to add a border around the quilt. How wide should he make the border to use all the fabric? (The border must be the same width on all four sides.)

Sketch the problem. Since you don’t know the width of the border, you will let the variable [latex]x[/latex] represent the width.

In the diagram, the original quilt is indicated by the red rectangle. The border is the area between the red and blue lines.

Since each side of the original 4 by 5 quilt has the border of width x added, the length of the quilt with the border will be [latex]5+2x[/latex], and the width will be [latex]4+2x[/latex].

(Both dimensions are written in terms of the same variable, and you will multiply them to get an area! This is where you might start to think that a quadratic equation might be used to solve this problem.)

You are only interested in the area of the border strips. Write an expression for the area of the border.

Area of border = Area of the blue rectangle minus the area of the red rectangle

Area of border[latex]=\left(4+2x\right)\left(5+2x\right)–\left(4\right)\left(5\right)[/latex]

There are 10 sq ft of fabric for the border, so set the area of border to be 10.

[latex]10=\left(4+2x\right)\left(5+2x\right)–20[/latex]

Multiply [latex]\left(4+2x\right)\left(5+2x\right)[/latex].

[latex]10=20+8x+10x+4x^{2}–20[/latex]

[latex]10=18x+4x^{2}[/latex]

Subtract 10 from both sides so that you have a quadratic equation in standard form and can apply the Quadratic Formula to find the roots of the equation.

[latex]\begin{array}{c}0=18x+4x^{2}-10\\\\\text{or}\\\\4x^{2}-10\\\\2\left(2x^{2}+9x-5\right)=0\end{array}[/latex]

Factor out the greatest common factor, 2, so that you can work with the simpler equivalent equation, [latex]2x^{2}+9x–5=0[/latex].

[latex]\large \begin{array}{r}2\left(2x^{2}+9x-5\right)=0\\\\\frac{2\left(2x^{2}+9x-5\right)}{2}=\frac{0}{2}\\\\2x^{2}+9x-5=0\end{array}[/latex]

Use the Quadratic Formula. In this case, [latex]a=2,b=9[/latex], and [latex]c=−5[/latex].

[latex]\large \begin{array}{l}x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\frac{-9\pm \sqrt{{{9}^{2}}-4(2)(-5)}}{2(2)}\end{array}[/latex]

[latex]\displaystyle x=\frac{-9\pm \sqrt{121}}{4}=\frac{-9\pm 11}{4}[/latex]

Find the solutions, making sure that the [latex]\pm[/latex] is evaluated for both values.

[latex]\large \begin{array}{c}x=\frac{-9+11}{4}=\frac{2}{4}=\frac{1}{2}=0.5\\\\\text{or}\\\\x=\frac{-9-11}{4}=\frac{-20}{4}=-5\end{array}[/latex]

Ignore the solution [latex]x=−5[/latex], since the width could not be negative.

The width of the border should be 0.5 ft.

Here is a video which gives another example of using the quadratic formula for a geometry problem involving the border around a quilt.

There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue.

Two graphs where the first graph shows the maximum value for f(x)=(x-2)^2+1 which occurs at (2, 1) and the second graph shows the minimum value for g(x)=-(x+3)^2+4 which occurs at (-3, 4).

Find two numbers [latex]x[/latex] and [latex]y[/latex] whose difference is 100 and whose product is a minimum.

We are trying to find the minimum of the product [latex]P=xy[/latex] of two numbers, such that their difference is 100: [latex]y-x=100[/latex]. First, we rewrite one variable in terms of the other:

[latex]y-x=100 \rightarrow y=100+x[/latex]

Next, we plug in the above relationship between the variables into the first equation:

[latex]P=xy=x(100+x) = 100x+x^2 = x^2+100x[/latex]

As a result, we get a quadratic function [latex]P(x)=x^2+100x[/latex]. The graph of this quadratic function opens upwards, and its vertex is the minimum, So if we find the vertex of this parabola, we will find the minimum product. The vertex is:

[latex]\displaystyle \displaystyle \left(-\frac{b}{2a}, P\left(-\frac{b}{2a}\right)\right) = \left(-\frac{(100)}{2(1)}, P\left(-\frac{(100)}{2(1)}\right)\right) = (-50,-2,500)[/latex]

Thus the minimum of the parabola occurs at [latex]x=-50[/latex], and is [latex]-2,500[/latex]. So one of the numbers is [latex]x=-50[/latex], the other we obtain by plugging in [latex]x=-50[/latex] in to [latex]y-x=100[/latex]:

[latex]\begin{array}{cc}y-(-50)&=&100 \\ y+50 &=& 100 \\ y &=& 50\end{array}[/latex]

[latex]x=-50[/latex] and [latex]y=50[/latex]

Example: Finding the Maximum Value of a Quadratic Function

A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.

Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length [latex]L[/latex].
What dimensions should she make her garden to maximize the enclosed area?

Let’s use a diagram such as the one above to record the given information. It is also helpful to introduce a temporary variable, W , to represent the width of the garden and the length of the fence section parallel to the backyard fence.

1) We know we have only 80 feet of fence available, and [latex]L+W+L=80[/latex], or more simply, [latex]2L+W=80[/latex]. This allows us to represent the width, [latex]W[/latex], in terms of [latex]L[/latex].

[latex]W=80 - 2L[/latex]

Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so

[latex]\begin{array}{l}\text{ }A&=&LW=L\left(80 - 2L\right)\hfill \\ A\left(L\right)&=&80L - 2{L}^{2}\hfill \end{array}[/latex]

This formula represents the area of the fence in terms of the variable length [latex]L[/latex]. The function, written in general form, is

[latex]A\left(L\right)=-2{L}^{2}+80L[/latex].

2) The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since [latex]a[/latex] is the coefficient of the squared term, [latex]a=-2,b=80[/latex], and [latex]c=0[/latex].

To find the vertex:

[latex]\large \begin{array}{l}h=-\frac{80}{2\left(-2\right)}\hfill & \hfill & \hfill & \hfill & k=A\left(20\right)\hfill \\ \text{ }=20\hfill & \hfill & \text{and}\hfill & \hfill & \text{ }=80\left(20\right)-2{\left(20\right)}^{2}\hfill \\ \hfill & \hfill & \hfill & \hfill & \text{ }=800\hfill \end{array}[/latex]

The maximum value of the function is an area of 800 square feet, which occurs when [latex]L=20[/latex] feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet.

Analysis of the Solution

This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function below.

Graph of the parabolic function A(L)=-2L^2+80L, which the x-axis is labeled Length (L) and the y-axis is labeled Area (A). The vertex is at (20, 800).

How To: Given an application involving revenue, use a quadratic equation to find the maximum.

Write a quadratic equation for revenue.
Find the vertex of the quadratic equation.
Determine the [latex]y[/latex]-value of the vertex.

Example: Finding Maximum Revenue

The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue?

Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, [latex]p[/latex] for price per subscription and [latex]Q[/latex] for quantity, giving us the equation [latex]\text{Revenue}=pQ[/latex].

Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently [latex]p=30[/latex] and [latex]Q=84,000[/latex]. We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, [latex]p=32[/latex] and [latex]Q=79,000[/latex]. From this we can find a linear equation relating the two quantities. The slope will be

[latex]\large \begin{array}{c}m=\frac{79,000 - 84,000}{32 - 30}\hfill \\ \text{ }=\frac{-5,000}{2}\hfill \\ \text{ }=-2,500\hfill \end{array}[/latex]

This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the y -intercept.

[latex]\begin{array}{c}\text{ }Q=-2500p+b\hfill & \text{Substitute in the point }Q=84,000\text{ and }p=30\hfill \\ 84,000=-2500\left(30\right)+b\hfill & \text{Solve for }b\hfill \\ \text{ }b=159,000\hfill & \hfill \end{array}[/latex]

This gives us the linear equation [latex]Q=-2,500p+159,000[/latex] relating cost and subscribers. We now return to our revenue equation.

[latex]\begin{array}{c}\text{Revenue}=pQ\hfill \\ \text{Revenue}=p\left(-2,500p+159,000\right)\hfill \\ \text{Revenue}=-2,500{p}^{2}+159,000p\hfill \end{array}[/latex]

We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex.

[latex]\large \begin{array}{c}h=-\frac{159,000}{2\left(-2,500\right)}\hfill \\ \text{ }=31.8\hfill \end{array}[/latex]

The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function.

[latex]\begin{array}{c}\text{maximum revenue}&=&-2,500{\left(31.8\right)}^{2}+159,000\left(31.8\right)\hfill \\ \text{ }&=&2,528,100\hfill \end{array}[/latex]

This could also be solved by graphing the quadratic. We can see the maximum revenue on a graph of the quadratic function.

Graph of the parabolic function which the x-axis is labeled Price (p) and the y-axis is labeled Revenue ($). The vertex is at (31.80, 258100).

A coordinate grid has been superimposed over the quadratic path of a basketball in the picture below. Find an equation for the path of the ball. Does the shooter make the basket?

Stop motioned picture of a boy throwing a basketball into a hoop to show the parabolic curve it makes.

(credit: modification of work by Dan Meyer)

The path passes through the origin and has vertex at [latex]\left(-4,\text{ }7\right)[/latex], so [latex]\left(h\right)x=-\frac{7}{16}{\left(x+4\right)}^{2}+7[/latex]. To make the shot, [latex]h\left(-7.5\right)[/latex] would need to be about 4 but [latex]h\left(-7.5\right)\approx 1.64[/latex]; he doesn’t make it.

Quadratic Formula Application - Time for an Object to Hit the Ground. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/RcVeuJhcuL0 . License : CC BY: Attribution
Quadratic Formula Application - Determine the Width of a Border. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/Zxe-SdwutxA . License : CC BY: Attribution
College Algebra. Authored by : Abramson, Jay, et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:1/Preface . License : CC BY: Attribution . License Terms : Download for free at : http://cnx.org/contents/[email protected]:1/Preface
Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution

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Chapter 10: Quadratics

10.7 Quadratic Word Problems: Age and Numbers

Quadratic-based word problems are the third type of word problems covered in MATQ 1099, with the first being linear equations of one variable and the second linear equations of two or more variables. Quadratic equations can be used in the same types of word problems as you encountered before, except that, in working through the given data, you will end up constructing a quadratic equation. To find the solution, you will be required to either factor the quadratic equation or use substitution.

Example 10.7.1

The sum of two numbers is 18, and the product of these two numbers is 56. What are the numbers?

First, we know two things:

[latex]\begin{array}{l} \text{smaller }(S)+\text{larger }(L)=18\Rightarrow L=18-S \\ \\ S\times L=56 \end{array}[/latex]

Substituting [latex]18-S[/latex] for [latex]L[/latex] in the second equation gives:

[latex]S(18-S)=56[/latex]

Multiplying this out gives:

[latex]18S-S^2=56[/latex]

Which rearranges to:

[latex]S^2-18S+56=0[/latex]

Second, factor this quadratic to get our solution:

[latex]\begin{array}{rrrrrrl} S^2&-&18S&+&56&=&0 \\ (S&-&4)(S&-&14)&=&0 \\ \\ &&&&S&=&4, 14 \end{array}[/latex]

[latex]\begin{array}{l} S=4, L=18-4=14 \\ \\ S=14, L=18-14=4 \text{ (this solution is rejected)} \end{array}[/latex]

Example 10.7.2

The difference of the squares of two consecutive even integers is 68. What are these numbers?

The variables used for two consecutive integers (either odd or even) is [latex]x[/latex] and [latex]x + 2[/latex]. The equation to use for this problem is [latex](x + 2)^2 - (x)^2 = 68[/latex]. Simplifying this yields:

[latex]\begin{array}{rrrrrrrrr} &&(x&+&2)^2&-&(x)^2&=&68 \\ x^2&+&4x&+&4&-&x^2&=&68 \\ &&&&4x&+&4&=&68 \\ &&&&&-&4&&-4 \\ \hline &&&&&&\dfrac{4x}{4}&=&\dfrac{64}{4} \\ \\ &&&&&&x&=&16 \end{array}[/latex]

This means that the two integers are 16 and 18.

Example 10.7.3

The product of the ages of Sally and Joey now is 175 more than the product of their ages 5 years prior. If Sally is 20 years older than Joey, what are their current ages?

The equations are:

[latex]\begin{array}{rrl} (S)(J)&=&175+(S-5)(J-5) \\ S&=&J+20 \end{array}[/latex]

Substituting for S gives us:

[latex]\begin{array}{rrrrrrrrcrr} (J&+&20)(J)&=&175&+&(J&+&20-5)(J&-&5) \\ J^2&+&20J&=&175&+&(J&+&15)(J&-&5) \\ J^2&+&20J&=&175&+&J^2&+&10J&-&75 \\ -J^2&-&10J&&&-&J^2&-&10J&& \\ \hline &&\dfrac{10J}{10}&=&\dfrac{100}{10} &&&&&& \\ \\ &&J&=&10 &&&&&& \end{array}[/latex]

This means that Joey is 10 years old and Sally is 30 years old.

For Questions 1 to 12, write and solve the equation describing the relationship.

The sum of two numbers is 22, and the product of these two numbers is 120. What are the numbers?
The difference of two numbers is 4, and the product of these two numbers is 140. What are the numbers?
The difference of two numbers is 8, and the sum of the squares of these two numbers are 320. What are the numbers?
The sum of the squares of two consecutive even integers is 244. What are these numbers?
The difference of the squares of two consecutive even integers is 60. What are these numbers?
The sum of the squares of two consecutive even integers is 452. What are these numbers?
Find three consecutive even integers such that the product of the first two is 38 more than the third integer.
Find three consecutive odd integers such that the product of the first two is 52 more than the third integer.
The product of the ages of Alan and Terry is 80 more than the product of their ages 4 years prior. If Alan is 4 years older than Terry, what are their current ages?
The product of the ages of Cally and Katy is 130 less than the product of their ages in 5 years. If Cally is 3 years older than Katy, what are their current ages?
The product of the ages of James and Susan in 5 years is 230 more than the product of their ages today. What are their ages if James is one year older than Susan?
The product of the ages (in days) of two newborn babies Simran and Jessie in two days will be 48 more than the product of their ages today. How old are the babies if Jessie is 2 days older than Simran?

Example 10.7.4

Doug went to a conference in a city 120 km away. On the way back, due to road construction, he had to drive 10 km/h slower, which resulted in the return trip taking 2 hours longer. How fast did he drive on the way to the conference?

The first equation is [latex]r(t) = 120[/latex], which means that [latex]r = \dfrac{120}{t}[/latex] or [latex]t = \dfrac{120}{r}[/latex].

For the second equation, [latex]r[/latex] is 10 km/h slower and [latex]t[/latex] is 2 hours longer. This means the second equation is [latex](r - 10)(t + 2) = 120[/latex].

We will eliminate the variable [latex]t[/latex] in the second equation by substitution:

[latex](r-10)(\dfrac{120}{r}+2)=120[/latex]

Multiply both sides by [latex]r[/latex] to eliminate the fraction, which leaves us with:

[latex](r-10)(120+2r)=120r[/latex]

Multiplying everything out gives us:

[latex]\begin{array}{rrrrrrrrr} 120r&+&2r^2&-&1200&-&20r&=&120r \\ &&2r^2&+&100r&-&1200&=&120r \\ &&&-&120r&&&&-120r \\ \hline &&2r^2&-&20r&-&1200&=&0 \end{array}[/latex]

This equation can be reduced by a common factor of 2, which leaves us with:

[latex]\begin{array}{rrl} r^2-10r-600&=&0 \\ (r-30)(r+20)&=&0 \\ r&=&30\text{ km/h or }-20\text{ km/h (reject)} \end{array}[/latex]

Example 10.7.5

Mark rows downstream for 30 km, then turns around and returns to his original location. The total trip took 8 hr. If the current flows at 2 km/h, how fast would Mark row in still water?

If we let [latex]t =[/latex] the time to row downstream, then the time to return is [latex]8\text{ h}- t[/latex].

The first equation is [latex](r + 2)t = 30[/latex]. The stream speeds up the boat, which means [latex]t = \dfrac{30}{(r + 2)}[/latex], and the second equation is [latex](r - 2)(8 - t) = 30[/latex] when the stream slows down the boat.

We will eliminate the variable [latex]t[/latex] in the second equation by substituting [latex]t=\dfrac{30}{(r+2)}[/latex]:

[latex](r-2)\left(8-\dfrac{30}{(r+2)}\right)=30[/latex]

Multiply both sides by [latex](r + 2)[/latex] to eliminate the fraction, which leaves us with:

[latex](r-2)(8(r+2)-30)=30(r+2)[/latex]

[latex]\begin{array}{rrrrrrrrrrr} (r&-&2)(8r&+&16&-&30)&=&30r&+&60 \\ &&(r&-&2)(8r&+&(-14))&=&30r&+&60 \\ 8r^2&-&14r&-&16r&+&28&=&30r&+&60 \\ &&8r^2&-&30r&+&28&=&30r&+&60 \\ &&&-&30r&-&60&&-30r&-&60 \\ \hline &&8r^2&-&60r&-&32&=&0&& \end{array}[/latex]

This equation can be reduced by a common factor of 4, which will leave us:

[latex]\begin{array}{rll} 2r^2-15r-8&=&0 \\ (2r+1)(r-8)&=&0 \\ r&=&-\dfrac{1}{2}\text{ km/h (reject) or }r=8\text{ km/h} \end{array}[/latex]

For Questions 13 to 20, write and solve the equation describing the relationship.

A train travelled 240 km at a certain speed. When the engine was replaced by an improved model, the speed was increased by 20 km/hr and the travel time for the trip was decreased by 1 hr. What was the rate of each engine?
Mr. Jones visits his grandmother, who lives 100 km away, on a regular basis. Recently, a new freeway has opened up, and although the freeway route is 120 km, he can drive 20 km/h faster on average and takes 30 minutes less time to make the trip. What is Mr. Jones’s rate on both the old route and on the freeway?
If a cyclist had travelled 5 km/h faster, she would have needed 1.5 hr less time to travel 150 km. Find the speed of the cyclist.
By going 15 km per hr faster, a transit bus would have required 1 hr less to travel 180 km. What was the average speed of this bus?
A cyclist rides to a cabin 72 km away up the valley and then returns in 9 hr. His speed returning is 12 km/h faster than his speed in going. Find his speed both going and returning.
A cyclist made a trip of 120 km and then returned in 7 hr. Returning, the rate increased 10 km/h. Find the speed of this cyclist travelling each way.
The distance between two bus stations is 240 km. If the speed of a bus increases by 36 km/h, the trip would take 1.5 hour less. What is the usual speed of the bus?
A pilot flew at a constant speed for 600 km. Returning the next day, the pilot flew against a headwind of 50 km/h to return to his starting point. If the plane was in the air for a total of 7 hours, what was the average speed of this plane?

Example 10.7.6

Find the length and width of a rectangle whose length is 5 cm longer than its width and whose area is 50 cm 2 .

First, the area of this rectangle is given by [latex]L\times W[/latex], meaning that, for this rectangle, [latex]L\times W=50[/latex], or [latex](W+5)W=50[/latex].

Multiplying this out gives us:

[latex]W^2+5W=50[/latex]

[latex]W^2+5W-50=0[/latex]

Second, we factor this quadratic to get our solution:

[latex]\begin{array}{rrrrrrl} W^2&+&5W&-&50&=&0 \\ (W&-&5)(W&+&10)&=&0 \\ &&&&W&=&5, -10 \\ \end{array}[/latex]

We reject the solution [latex]W = -10[/latex].

This means that [latex]L = W + 5 = 5+5= 10[/latex].

Example 10.7.7

If the length of each side of a square is increased by 6, the area is multiplied by 16. Find the length of one side of the original square.

The relationship between these two is:

[latex]\begin{array}{rrl} \text{larger area}&=&16\text{ times the smaller area} \\ (x+12)^2&=&16(x)^2 \end{array}[/latex]

Simplifying this yields:

[latex]\begin{array}{rrrrrrr} x^2&+&24x&+&144&=&16x^2 \\ -16x^2&&&&&&-16x^2 \\ \hline -15x^2&+&24x&+&144&=&0 \end{array}[/latex]

Since this is a problem that requires factoring, it is easiest to use the quadratic equation:

[latex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a},\hspace{0.25in}\text{ where }a=-15, b=24\text{ and }c=144[/latex]

Substituting these values in yields [latex]x = 4[/latex] or [latex]x=-2.4[/latex] (reject).

Example 10.7.8

Nick and Chloe want to surround their 60 by 80 cm wedding photo with matting of equal width. The resulting photo and matting is to be covered by a 1 m 2 sheet of expensive archival glass. Find the width of the matting.

[latex](L+2x)(W+2x)=1\text{ m}^2[/latex]

[latex](80\text{ cm }+2x)(60\text{ cm }+2x)=10,000\text{ cm}^2[/latex]

[latex]4800+280x+4x^2=10,000[/latex]

[latex]4x^2+280x-5200=0[/latex]

Which reduces to:

[latex]x^2 + 70x - 1300 = 0[/latex]

Second, we factor this quadratic to get our solution.

It is easiest to use the quadratic equation to find our solutions.

[latex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a},\hspace{0.25in}\text{ where }a=1, b=70\text{ and }c=-1300[/latex]

Substituting the values in yields:

[latex]x=\dfrac{-70\pm \sqrt{70^2-4(1)(-1300)}}{2(1)}\hspace{0.5in}x=\dfrac{-70\pm 10\sqrt{101}}{2}[/latex]

[latex]x=-35+5\sqrt{101}\hspace{0.75in} x=-35-5\sqrt{101}\text{ (rejected)}[/latex]

For Questions 21 to 28, write and solve the equation describing the relationship.

Find the length and width of a rectangle whose length is 4 cm longer than its width and whose area is 60 cm 2 .
Find the length and width of a rectangle whose width is 10 cm shorter than its length and whose area is 200 cm 2 .
A large rectangular garden in a park is 120 m wide and 150 m long. A contractor is called in to add a brick walkway to surround this garden. If the area of the walkway is 2800 m 2 , how wide is the walkway?
A park swimming pool is 10 m wide and 25 m long. A pool cover is purchased to cover the pool, overlapping all 4 sides by the same width. If the covered area outside the pool is 74 m 2 , how wide is the overlap area?
In a landscape plan, a rectangular flowerbed is designed to be 4 m longer than it is wide. If 60 m 2 are needed for the plants in the bed, what should the dimensions of the rectangular bed be?
If the side of a square is increased by 5 units, the area is increased by 4 square units. Find the length of the sides of the original square.
A rectangular lot is 20 m longer than it is wide and its area is 2400 m 2 . Find the dimensions of the lot.
The length of a room is 8 m greater than its width. If both the length and the width are increased by 2 m, the area increases by 60 m 2 . Find the dimensions of the room.

Answer Key 10.7

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Procedure Steps, Example Solved Problem - Solving Problems Involving Quadratic Equations | 10th Mathematics : UNIT 3 : Algebra

Chapter: 10th mathematics : unit 3 : algebra.

Solving Problems Involving Quadratic Equations

Steps to solve a problem

Step 1 Convert the word problem to a quadratic equation form

Step 2 Solve the quadratic equation obtained in any one of the above three methods.

Step 3 Relate the mathematical solution obtained to the statement asked in the question.

Example 3.37

The product of Kumaran’s age (in years) two years ago and his age four years from now is one more than twice his present age. What is his present age?

Let the present age of Kumaran be x years.

Two years ago, his age = ( x − 2) years.

Four years from now, his age = ( x + 4) years.

( x − 2)( x + 4) = 1 +2 x

x 2 + 2 x − 8 = 1 +2 x gives ( x − 3)( x + 3) = 0 then, x = ± 3

Therefore, x = 3 (Rejecting −3 as age cannot be negative)

Kumaran’s present age is 3 years.

Example 3.38

A ladder 17 feet long is leaning against a wall. If the ladder, vertical wall and the floor from the bottom of the wall to the ladder form a right triangle, find the height of the wall where the top of the ladder meets if the distance between bottom of the wall to bottom of the ladder is 7 feet less than the height of the wall?

Let the height of the wall AB = x feet

As per the given data BC = ( x –7) feet

In the right triangle ABC , AC =17 ft, BC = ( x –7) feet

By Pythagoras theorem, AC 2 = AB 2 + BC 2

(17) 2 = x 2 + ( x − 7) 2 ; 289 = x 2 + x 2 − 14 x + 49

x 2 − 7 x −120 = 0 hence, ( x − 15)( x + 8) = 0 then, x = 15 (or) −8

Therefore, height of the wall AB = 15 ft (Rejecting −8 as height cannot be negative)

Example 3.39

A flock of swans contained x 2 members. As the clouds gathered, 10 x went to a lake and one-eighth of the members flew away to a garden. The remaining three pairs played about in the water. How many swans were there in total?

As given there are x 2 swans.

As per the given data x 2 − 10 x – (1/8) x 2 = 6 we get, 7 x 2 − 680 x − 48 = 0

Therefore, x = 12, -4/7

Here x = 4/7 is not possible as the number of swans cannot be negative.

Hence, x = 12. Therefore total number of swans is x 2 = 144.

Example 3.40

A passenger train takes 1 hr more than an express train to travel a distance of 240 km from Chennai to Virudhachalam. The speed of passenger train is less than that of an express train by 20 km per hour. Find the average speed of both the trains.

Let the average speed of passenger train be x km/hr.

Then the average speed of express train will be ( x + 20) km/hr

Time taken by the passenger train to cover distance of 240 km = 240/ x hr

Time taken by express train to cover distance of 240 km = 240 / ( x +20) hr

x 2 + 20 x – 4800 = 0 gives, ( x + 80)( x − 60) = 0 we get, x = –80 or 60.

Therefore x = 60 (Rejecting -80 as speed cannot be negative)

Average speed of the passenger train is 60 km/hr

Average speed of the express train is 80 km/hr.

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Mathematics 9 Quarter 1-Module 9: Solving Problems Involving Quadratic Equations (Week 4 Learning Code – M9AL-1e-1)

In the previous lessons, you learned about the different ways in solving quadratic equations, the concepts of the “nature of its roots” and the relationship of its roots and coefficients. Your mastery of the lessons is an important tool to solve many real-world problems related to quadratic equations. In this module, you will learn how to solve word problems that involves quadratic equations.

LEARNING COMPETENCY

The learners will be able to:

solve problems involving quadratic equations. M9AL-1e-1

Can't Find What You'RE Looking For?

We are here to help - please use the search box below.

SOLVING WORD PROBLEMS INVOLVING QUADRATIC EQUATIONS

Problem 1 :

If the difference between a number and its reciprocal is ²⁴⁄₅ , find the number.

Let x be the required number. Then its reciprocal is ¹⁄ₓ be its reciprocal.

5 (x 2 - 1) = 24x

5 x 2 - 5 = 24x

5x 2 - 24x - 5 = 0

5x 2 - 25x + 1x - 5 = 0

5x(x - 5) + 1(x - 5) = 0

(5x + 1)(x - 5) = 0

5x + 1 = 0 or x - 5 = 0

x = - ⅕ or x = 5

So, the required number is - ⅕ or -5.

Problem 2 :

A garden measuring 12m by 16m is to have a pedestrian pathway that is w meters wide installed all the way around so that it increases the total area to 285 m 2 . What is the width of the pathway?

From the picture above, length of the garden including pathway is (12 + 2w) and width is (16 + 2w).

Total Area = 285 m 2

Length ⋅ Width = 285

(12 + 2w) (16 + 2w) = 285

192 + 24w + 32w + 4w 2 = 285

4w 2 + 56w + 192 - 285 = 0

4w 2 + 56w - 93 = 0

Solve the above quadratic equation using quadratic formula.

w = -15.5 or w = 1.5

Since w is the width of the pathway, it can not be negative. So, w = 1.5.

Therefore , the width of the pathway is 1.5 m .

Problem 3 :

A bus covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the bus.

Distance covered = 90 km

Let x be the original speed of the bus.

Increased speed = x + 15

Time taken by the bus in original speed :

Time taken by the bus in increased speed :

= ⁹⁰⁄₍ₓ ₊ ₁₅₎

From the given information, the difference between and ⁹⁰⁄ₓ and ⁹⁰⁄₍ₓ ₊ ₁₅₎ is 30 minutes or ½ hour .

2700 = x 2 + 15x

x 2 + 15x - 2700 = 0

x 2 + 60x - 45x - 2700 = 0

x(x + 60) - 45(x + 60) = 0

(x - 45)(x + 60) = 0

x = 45 or x = -60

Since x represents the original speed of the bus, it can never be negative. So, x = 45.

Therefore, the original speed of the bus is 45 km per hour.

Problem 4 :

John and Jivanti together had 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they have now is 124. Find the number marbles each one them had initially.

Let x be the number of marbles that john has.

Then, the number of marbles that Jivanthi has

After losing 5 marbles,

number of marbles with John = x - 5

number of marbles with Jivanti = 45 - x - 5 = 40 - x

The product of number of marbles = 124

(x - 5)(40 - x) = 124

40x - x 2 - 200 + 5x = 124

45x - x 2 - 200 = 124

x 2 - 45x + 124 + 200 = 0

x 2 - 45x + 324 = 0

x 2 - 36x - 9x + 324 = 0

x(x - 36) - 9(x - 36) = 0

(x - 9)(x - 36) = 0

x - 9 = 0 or x - 36 = 0

x = 9 or x = 36

If x = 9,

If x = 36,

So, John had 36 marbles, when Jivanti had 9 marbles or Jivanti had 36 marbles, when John had 9 marbles.

Problem 5 :

A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in dollars) was found to be 55 minus the number of toys produced in a day, the total cost of production was $750. Find the number of toys produced on that day.

Let x be the number of toys produced in a particular day.

Cost of production of one toy (in dollars) :

Total cost = Number of toys ⋅ Cost of one toy

750 = x (55 - x)

750 = 55x - x 2

x 2 - 55x + 750 = 0

x 2 - 30x - 25x + 750 = 0

x(x - 30) - 25(x - 30) = 0

(x - 30)(x - 25) = 0

x - 30 = 0 or x - 25 = 0

x = 30 or x = 25

So, the number of toys produced on that particular day is 30 or 25.

Problem 6 :

The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

Let x be the length of shorter side of the rectangle

length of diagonal = x + 60

length of longer side = x + 30

Using Pythagorean Theorem in the right triangle ABC,

(x + 60) 2 = x 2 + (x + 30 ) 2

x 2 + 60 2 + 2(x)(60) = x 2 + x 2 + 2(x)(30) + 30 2

x 2 + 3600 + 120x = 2x 2 + 60x + 900

2 x 2 - x 2 + 60x - 120x + 900 - 3600 = 0

x 2 - 60x - 2700 = 0

x 2 - 90x + 30x - 2700 = 0

x(x - 90) + 30(x - 90) = 0

(x + 30)(x - 90) = 0

x + 30 = 0 or x - 90 = 0

x = -30 or x = 90

Because x represents breadth of the rectangle, it can never be negative. So, x = 90.

Therefore,

length of the shorter side = 90 m

length of the longer side = 90 + 30 = 120 m

Problem 7 :

The difference of squares of two positive numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Let x be the larger number and y be the smaller number.

The square of the smaller number is 8 times the larger number.

y 2 = 8x -----(1)

The difference of squares of two numbers is 180.

x 2 - y 2 = 180

Substitute y 2 = 8x and solve for x.

x 2 - 8x = 180

x 2 - 8x - 180 = 0

x 2 - 18x + 10x - 180 = 0

x(x - 18) + 10(x - 18) = 0

(x - 18)(x + 10) = 0

x - 18 = 0 or x + 10 = 0

x = 18 or x = -10

Because the numbers are positive, x = -10 can not be accepted. So, x = 18.

Substitute x = 18 into (1).

y 2 = 8(18)

y 2 = 144

y = √144

y = 12

Therefore, the larger number is 18 and the smaller number is 12.

Problem 8 :

A train travels 360 miles at a uniform speed. If the speed had been 5 miles/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Distance covered = 360 miles

Let x be the original speed of the train.

If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey.

Increased speed = x + 5

Time taken by the train in original speed :

= ³⁶⁰⁄ₓ

Time taken by the train in increased speed :

= ³⁶⁰⁄₍ₓ ₊ ₅₎

From the given information, the difference between and ³⁶⁰⁄ₓ and ³⁶⁰⁄₍ₓ ₊ ₅₎ is 1 hour .

1800 = x 2 + 5x

x 2 + 5x - 1800 = 0

x 2 - 40x + 45x - 1800 = 0

x(x - 40) + 45(x - 40) = 0

(x - 40)(x + 45) = 0

x - 40 = 0 or x + 45 = 0

x = 40 or x = -45

Since x represents the original speed of the train, it can never be negative. So, x = 40.

Therefore, the original speed of the train is 40 miles/hr.

Problem 9 :

Find two consecutive positive even integers whose squares have the sum 340.

Let x and ( x + 2) be the two positive even integers.

Sum of their squares = 340

x 2 + (x + 2) 2 = 340

x 2 + (x + 2)(x + 2) = 340

x 2 + x 2 + 2x + 2x + 4 = 340

2x 2 + 4x + 4 = 340

2x 2 + 4x - 336 = 0

Divide both sides by 2.

x 2 + 2x - 168 = 0

Solve by factoring.

x 2 - 12x + 14x - 168 = 0

x(x - 12) + 14(x - 12) = 0

(x - 12)(x + 14) = 0

x - 12 = 0 or x + 14 = 0

x = 12 or x = -14

Since the integers are positive, x can not be negative.

Therefore, the two positive even integers are 12 and 14.

Problem 10 :

The sum of squares of three consecutive natural numbers is 194. Determine the numbers.

Let x , x + 1 and x + 2 be three consecutive natural numbers.

Sum of their squares = 194

x 2 + (x + 1) 2 + (x + 2) 2 = 194

x 2 + (x + 1)(x + 1) + (x + 2)(x + 2) = 194

x 2 + x 2 + x + x + 1 + x 2 + 2x + 2x + 4 = 194

3x 2 + 6x + 5 = 194

3x 2 + 6x - 189 = 0

Divide both sides by 3.

x 2 + 2x - 63 = 0

x 2 - 7x + 9x - 63 = 0

x(x - 7) + 9(x - 7) = 0

(x - 7)(x + 9) = 0

x - 7 = 0 or x + 9 = 0

x = 7 or x = -9

Since the numbers are natural numbers, x can not be negative.

Therefore, the three consecutive natural numbers are 7, 8 and 9.

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Incoming first-time students will receive a link to a personalized survey that builds their mathematical history. The survey directs them to the most appropriate placement test based on this history and their major of interest.

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Topics covered on each exam can be found below.

The placement test has 25 questions, a mix of multiple-choice and open-response. A score of 17 or higher satisfies placement into Calculus (MAT 128 or MAT 129). Topics covered on the exam include:

Simplify algebraic expressions involving roots and rational exponents.
Add, subtract, and multiply polynomials.
Factor a polynomial in one or more variables by factoring out its greatest common factor. Factor a trinomial. Factor the difference of squares.
Add, subtract, multiply, divide, and simplify rational expressions.
Solve linear and quadratic equations. Solve equations involving a radical, a rational, or an absolute value expression.
Solve linear and nonlinear inequalities in one variable.
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Solve problems that can be modeled using a linear or quadratic equation or expression.
Solve geometry problems using the Pythagorean theorem, the properties of similar triangles, and right-triangle trigonometry.
Understand and apply the relationship between the properties of a graph of a line and its equation.
Find the intercepts and the graph of a parabola given its equation. Find an equation of a parabola given its graph.
Find the domain of a function. Evaluate a function at a point in its domain.
Combine functions by composition, addition, subtraction, multiplication, and division. Find the inverse of a one-to-one function.
Understand the relationship between radian measure and the arc length of a circle.
Know values of the sine, cosine, and tangent functions as functions on the unit circle that are multiples of and . Sketch the graphs of these functions.
Apply basic trigonometric identities to solve problems and verify more complex identities. Solve simple trigonometric equations.
Understand and apply the properties of exponential and logarithmic functions.
Know that the inverse of an exponential function is a logarithm and use this inverse relationship to solve simple logarithmic and exponential equations.

The test has 25 multiple-choice questions. A score of 17 or higher satisfies placement into pre-calculus (MAT 119). Topics on the placement test include:

Evaluate algebraic expressions at specified values of their variables.
Simplify algebraic expressions involving square roots and cube roots.
Factor polynomials by factoring out the greatest common factor and factoring quadratic polynomials.
Add, subtract, multiply, and simplify rational expressions.
Solve linear equations and linear inequalities.
Solve a multivariable equation for one of its variables.
Solve quadratic equations by factoring and by using the quadratic formula.
Solve an equation involving a radical, a rational, or an absolute value expression.
Solve geometric problems using the Pythagorean theorem and the properties of similar triangles.
Evaluate a function at a number in its domain.
Find the domain of a rational function and the square root of a linear function.

The placement test has 25 multiple-choice questions. A score of 17 or higher satisfies placement into algebra (MAT 102). Topics on the placement test include:

Evaluate numerical expressions with signed numbers, exponents, and parentheses using order of operations.
Perform arithmetic operations with fractions, decimals, and percents.
Order fractions and decimals on a number line.
Solve applied problems using appropriate units, including problems involving percentage increase and decrease, rates, and proportions.
Solve geometric problems about rectangles and triangles.
Simplify algebraic expressions involving integer exponents and square roots.
Factor polynomials by factoring out the greatest common factor and factoring quadratic polynomials (with a leading coefficient of 1).
Use a linear equation to set up and solve a word problem.
Find an equation of a line given two points on the line, one point and the slope, or the graph of the line.
Given an equation of a line, find the slope and graph the line.

The placement test has 25 multiple-choice questions. A score of 17 or higher satisfies placement into general education coursework like STA 205 and MAT 115. Topics on the placement test include:

Solve simple coordinate geometry problems.
Solve a linear equation.
Use a linear equation to solve a simple word problem.

For more information regarding math placement needs, students are encouraged to reach out to their Academic Advisor.

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IMAGES

Quadratic Word Problems Worksheet
Problem solving with quadratic equations questions
Problem solving with quadratic equations questions
👍 Quadratic equation problem solving. General Quadratic Word Problems. 2019-02-13
👍 Quadratic equation problem solving. General Quadratic Word Problems. 2019-02-13
😀 Quadratic problem solving. Quadratic equations and problem solving. 2019-02-16

VIDEO

10th Maths
Solving Problems Involving Quadratic Equations
Solving Problems Involving Quadratic Equations
10th Maths
SOLVING PROBLEMS INVOLVING QUADRATIC EQUATIONS
indices involving quadratic equation

COMMENTS

9.6: Solve Applications of Quadratic Equations
Step 5: Solve the equation. Substitute in the values. Distribute. This is a quadratic equation; rewrite it in standard form. Solve the equation using the Quadratic Formula. Identify the $a,b,c$ values. Write the Quadratic Formula. Then substitute in the values of $a,b,c$. Simplify. Figure 9.5.13: Rewrite to show two solutions.
Word Problems Involving Quadratic Equations
A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h (t)= -16t 2 + 40ft + 1.5. Find the maximum height attained by the ball. Let's first take a minute to understand this problem and what it means. We know that a ball is being shot from a cannon.
Quadratic functions & equations
Solve by completing the square: Non-integer solutions. Worked example: completing the square (leading coefficient ≠ 1) Solving quadratics by completing the square: no solution. Proof of the quadratic formula. Solving quadratics by completing the square. Completing the square review. Quadratic formula proof review.
Quadratic equations word problem
Quadratic word problem: ball. Quadratic word problems (standard form) Math > Algebra 1 > Quadratic functions & equations > Quadratic standard form ... So we're really solving the equation 0 is equal to negative 16t squared plus 20t plus 50. And if you want to simplify this a little bit-- let's see, everything here is divisible at least by 2. ...
Quadratic equations & functions
Solving quadratics by completing the square. Worked example: Completing the square (intro) Worked example: Rewriting expressions by completing the square. Worked example: Rewriting & solving equations by completing the square. Worked example: completing the square (leading coefficient ≠ 1) Solving quadratics by completing the square: no solution.
9.6: Solve Applications of Quadratic Equations
Step 5. Solve the equation using algebra techniques. Step 6. Check the answer in the problem and make sure it makes sense. Step 7. Answer the question with a complete sentence. We have solved number applications that involved consecutive even and odd integers, by modeling the situation with linear equations.
10 Quadratic Equations Word Problems
Quadratic equations word problems are math problems in which the equations are not given directly. These problems can be solved by using the given information to obtain a quadratic equation of the form ax^2+bx+c ax2 + bx+ c. We can then use the factoring method, the completing the square method or the quadratic formula to solve the equation.
Real World Examples of Quadratic Equations
Step 1 Divide all terms by -200. P 2 - 460P + 42000 = 0. Step 2 Move the number term to the right side of the equation: P 2 - 460P = -42000. Step 3 Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation: (b/2) 2 = (−460/2) 2 = (−230) 2 = 52900.
Quadratic Equation Calculator
In math, a quadratic equation is a second-order polynomial equation in a single variable. It is written in the form: ax^2 + bx + c = 0 where x is the variable, and a, b, and c are constants, a ≠ 0.
10.3
(10.3.1) - Solve application problems involving quadratic functions. Quadratic equations are widely used in science, business, and engineering. Quadratic equations are commonly used in situations where two things are multiplied together and they both depend on the same variable.
SOLVING PROBLEMS INVOLVING QUADRATIC EQUATION
For more examples pls. watch this link.https://youtu.be/rLBHhHId52oGrade 9 LessonSOLVING PROBLEMS INVOLVING QUADRATIC EQUATION00:00 - Intro03:29 - Number Sen...
Quadratic formula explained (article)
First we need to identify the values for a, b, and c (the coefficients). First step, make sure the equation is in the format from above, a x 2 + b x + c = 0 : x 2 + 4 x − 21 = 0. is what makes it a quadratic). Then we plug a , b , and c into the formula: x = − 4 ± 16 − 4 ⋅ 1 ⋅ ( − 21) 2. solving this looks like:
Quadratic Equation Solver
There are different methods you can use to solve quadratic equations, depending on your particular problem. Solve By Factoring. Example: 3x^2-2x-1=0. Complete The Square. Example: 3x^2-2x-1=0 (After you click the example, change the Method to 'Solve By Completing the Square'.) Take the Square Root. Example: 2x^2=18. Quadratic Formula
10.7 Quadratic Word Problems: Age and Numbers
Quadratic-based word problems are the third type of word problems covered in MATQ 1099, with the first being linear equations of one variable and the second linear equations of two or more variables. Quadratic equations can be used in the same types of word problems as you encountered before, except that, in working through the given data, you ...
Solving Problems Involving Quadratic Equations
Solving Problems Involving Quadratic Equations. Steps to solve a problem. Step 1 Convert the word problem to a quadratic equation form. Step 2 Solve the quadratic equation obtained in any one of the above three methods.. Step 3 Relate the mathematical solution obtained to the statement asked in the question.. Example 3.37. The product of Kumaran's age (in years) two years ago and his age ...
Solving Quadratic Equations
Methods of solving quadratic equations are discussed here in the following steps. Step I: Denote the unknown quantities by x, y etc. Step II: use the conditions of the problem to establish in unknown quantities. Step III: Use the equations to establish one quadratic equation in one unknown. Step IV: Solve this equation to obtain the value of ...
PDF GCSE TOPIC BOOKLET PROBLEMS INVOLVING QUADRATIC EQUATIONS
In the triangle below, show that x satisfies the equation 3x2 — o. -3x cm [3] 300 Solve the equation 3x2 — place. — 2) cm Diagram not drawn to scale — O and hence find the length of BC, correct to 1 decimal [4]
Solving quadratic equations
When solving quadratic equations by taking square roots, both the positive and negative square roots are solutions to the equation. This is because when we square a solution, the result is always positive. For example, for the equation x 2 = 4 , both 2 and − 2 are solutions: 2 2 = 4. ‍.
Mathematics 9 Quarter 1-Module 9: Solving Problems Involving Quadratic
Your mastery of the lessons is an important tool to solve many real-world problems related to quadratic equations. In this module, you will learn how to solve word problems that involves quadratic equations. LEARNING COMPETENCY. The learners will be able to: solve problems involving quadratic equations. M9AL-1e-1; Math9_Quarter1_Module9_FINAL-V3-1
SOLVING WORD PROBLEMS INVOLVING QUADRATIC EQUATIONS
Solve the above quadratic equation using quadratic formula. w = -15.5 or w = 1.5 Since w is the width of the pathway, it can not be negative. So, w = 1.5. Therefore , the width of the pathway is 1.5 m. Problem 3 : A bus covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more, it would have taken 30 minutes less for the ...
Quadratic word problems (standard form) (practice)
Quadratic word problems (standard form) Google Classroom. You might need: Calculator. Rui is a professional deep water free diver. His altitude (in meters relative to sea level), x seconds after diving, is modeled by: d ( x) = 1 2 x 2 − 10 x.
Math Placement: Northern Kentucky University, Greater Cincinnati Region
Solve quadratic equations by factoring and by using the quadratic formula. Solve an equation involving a radical, a rational, or an absolute value expression. Solve a system of two linear equations in two variables. Solve problems that can be modeled using a linear or quadratic equation or expression. Solve geometric problems using the ...