 Section 6.4 Quadratic Formula
Practice Solutions
Corrective Assignment
Application Walkthrough
Quadratic Formula ExercisesQuadratic formula practice problems with answers. Below are ten (10) practice problems regarding the quadratic formula. The more you use the formula to solve quadratic equations, the more you become expert at it! Use the illustration below as a guide. Notice that in order to apply the quadratic formula, we must transform the quadratic equation into the standard form, that is, [latex]a{x^2} + bx + c = 0[/latex] where [latex]a \ne 0[/latex]. The problems below have varying levels of difficulty. I encourage you to try them all. Believe me, they are actually easy! Good luck.  Problem 1: Solve the quadratic equation using the quadratic formula. [latex]{x^2}\, – \,8x + 12 = 0[/latex] Therefore, the answers are [latex]{x_1} = 6[/latex] and [latex]{x_2} = 2[/latex]. Problem 2: Solve the quadratic equation using the quadratic formula. [latex]2{x^2}\, -\, x = 1[/latex] Rewrite the quadratic equation in the standard form. [latex]2{x^2} – x – 1 = 0[/latex] Therefore, the answers are [latex]{x_1} = 1[/latex] and [latex]{x_2} = \large{{ – 1} \over 2}[/latex]. Problem 3: Solve the quadratic equation using the quadratic formula. [latex]4{x^2} + 9 = – 12x[/latex] [latex]4{x^2} + 12x + 9 = 0[/latex] Therefore, the solution is [latex]x = \large{{ – 3} \over 2}[/latex]. Problem 4: Solve the quadratic equation using the quadratic formula. [latex]5{x^2} = 7x + 6[/latex] Convert the quadratic equation into the standard form. [latex]5{x^2} – 7x – 6 = 0[/latex] Therefore, the answers are [latex]{x_1} = 2[/latex] and [latex]{x_2} = \large{{ – 3} \over 5}[/latex]. Problem 5: Solve the quadratic equation using the quadratic formula. [latex]{x^2} -\,{ \large{1 \over 2}}x\, – \,{\large{3 \over {16}}} = 0[/latex] Multiply the entire equation by the LCM of the denominators which is [latex]16[/latex]. This will get rid of the denominators thereby giving us integer values for [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex]. [latex]16{x^2} – 8x – 3 = 0[/latex] Therefore, the answers are [latex]x_1=\large{3 \over 4}[/latex] and [latex]x_2=\large{{ – 1} \over 4}[/latex]. Problem 6: Solve the quadratic equation using the quadratic formula. [latex]{x^2} + 3x + 9 = 5x – 8[/latex] Convert into standard form as [latex]{x^2} – 2x + 17 = 0[/latex]. Therefore, the answers are [latex]x_1=1 + 4i[/latex] and [latex]x_2=1 – 4i[/latex]. Problem 7: Solve the quadratic equation using the quadratic formula. [latex]{\left( {x – 2} \right)^2} = 4x[/latex] Rewrite in standard form as [latex]{x^2} – 8x + 4 = 0[/latex]. Hence, the answers are [latex]{x_1} = 4 + 2\sqrt 3 [/latex] and [latex]{x_2} = 4 – 2\sqrt 3 [/latex]. Problem 8: Solve the quadratic equation using the quadratic formula. [latex]{\Large{{{x^2}} \over 4} – {x \over 2} }= 1[/latex] To convert the quadratic equation into the standard form, simply multiply the entire equation by [latex]4[/latex] then subtract both sides by [latex]4[/latex]. [latex]{x^2} – 2x – 4 = 0[/latex] Thus, the answers are [latex]{x_1} = 1 + \sqrt 5 [/latex] and [latex]{x_2} = 1 – \sqrt 5 [/latex]. Problem 9: Solve the quadratic equation using the quadratic formula. [latex]{\left( {2x – 1} \right)^2} = \Large{x \over 3}[/latex] If we carefully transform the given quadratic equation into the standard form, we get [latex]12{x^2} – 13x + 3 = 0[/latex]. Therefore, the answers are [latex]x_1={\Large{3 \over 4}}[/latex] and [latex]x_2={\Large{1 \over 3}}[/latex]. Problem 10: Solve the quadratic equation using the quadratic formula. [latex]\left( {2x – 1} \right)\left( {x + 4} \right) = – {x^2} + 3x[/latex] If we simplify the quadratic equation to convert it to the standard form, we should arrive at [latex]3{x^2} + 4x – 4 = 0[/latex]. Hence, the answers are [latex]x_1={\Large{2 \over 3}}[/latex] and [latex]x_2=-2[/latex]. You might also like these tutorials:
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Catch-Up and ReviewHere are a few recommended readings before getting started with this lesson.
Try a few practice exercises as a warm-up! Is There a Solution?Magdalena and Diego, both huge fans of statistics, went camping to bond under the stars and talk stats. However, they realize that bears are in the area. They need to hang their food basket from a branch 1 5 feet above the ground. Diego figures he can throw a stone with a rope attached to it over the branch. As Diego winds up, Magdalena sheepishly snickers, "No way that works." The Quadratic FormulaBesides graphing , using square roots , factoring , and completing the square , there is another method for solving a quadratic equation . This method consists of using the Quadratic Formula . Check out how to derive the formula by completing the square! Quadratic FormulaThe Quadratic Formula can be used to solve a quadratic equation written in standard form a x 2 + b x + c = 0 . x = 2 a - b ± b 2 − 4 a c b a / c = b ⋅ c a 2 ⋅ 2 a = a Commutative Property of Multiplication a 2 + 2 a b + b 2 = ( a + b ) 2 Commutative Property of Addition ( b a ) m = b m a m ( a b ) m = a m b m b a = b ⋅ 4 a a ⋅ 4 a a ⋅ a = a 2 Subtract fractions b a = b a a ⋅ b = a ⋅ b LHS − 2 a b = RHS − 2 a b Put minus sign in numerator Add and subtract fractions Solving a Quadratic Equation Using the Quadratic FormulaSince the profit should be at least $ 2 0 0 , let p ( x ) be equal to 2 0 0 . Then, rewrite the quadratic equation in standard form . The equation can be solved using the Quadratic Formula . p ( x ) = 2 0 0 LHS − 2 0 0 = RHS − 2 0 0 Rearrange equation Substitute values Calculate power a ( - b ) = - a ⋅ b ( - a ) ( - b ) = a ⋅ b Subtract term Calculate root
Since Magdalena wants the tickets to be as cheap as possible while making a profit of at least $ 2 0 0 , the price each ticket should be $ 4 . Solving a Quadratic Equation Not in Standard FormA fire nozzle attached to a hose is a device used by firefighters to extinguish fires. Consider a firefighter who is aiming water to extinguish a fire on the third floor of a building. The base of the fire is situated 2 2 feet above the ground. What is the height of the water stream's peak? Write a quadratic equation and solve it using the Quadratic Formula . Distribute - 0 . 0 0 8 x LHS − 2 4 = RHS − 2 4 Add and subtract terms - b - a = b a Calculate quotient Solve the quadratic equations by using the Quadratic Formula . If necessary, round the answer to 2 decimal places. Discriminant of a Quadratic EquationIn general, quadratic equations have two, one or no real solutions. Before solving a quadratic equation, the number of real solutions can be determined by using the discriminant . DiscriminantIn the Quadratic Formula , the expression b 2 − 4 a c , which is under the radical symbol, is called the discriminant . A quadratic equation can have two, one, or no real solutions. Since the discriminant is under the radical symbol, its value determines the number of real solutions of a quadratic equation.
Moreover, the discriminant determines the number of x - intercepts of the graph of the related quadratic function . Determining Whether There is a Solution Without SolvingLet x denote the length of one side of the rectangle. Then, use the fact that the length of the fence represents the perimeter of the rectangle. All things considered, how can the area of the rectangle be calculated? LHS − 2 x = RHS − 2 x LHS / 2 = RHS / 2 Write as a difference of fractions c a ⋅ b = c a ⋅ b Identity Property of Multiplication Distribute x LHS − 5 0 0 0 0 = RHS − 5 0 0 0 0 - a ( - b ) = a ⋅ b Determining the Number of Real Solutions of a Quadratic EquationWithout solving the quadratic equations , use the discriminant to determine the number of real solutions. Determining the Number of Solutions of a Quadratic Equation Without SolvingThe challenge presented at the beginning of this lesson asked if the stone thrown by Diego will reach, over some point in time, a branch located 1 5 feet above the ground. Substitute 1 5 for h ( t ) and identify the discriminant of the resulting quadratic equation . Distribute 5 LHS − 1 5 = RHS − 1 5 10.3 Solve Quadratic Equations Using the Quadratic FormulaLearning objectives. By the end of this section, you will be able to:
Be Prepared 10.7Before you get started, take this readiness quiz. Simplify: −20 − 5 10 −20 − 5 10 . If you missed this problem, review Example 1.74 . Be Prepared 10.8Simplify: 4 + 121 4 + 121 . If you missed this problem, review Example 9.29 . Be Prepared 10.9Simplify: 128 128 . If you missed this problem, review Example 9.12 . When we solved quadratic equations in the last section by completing the square, we took the same steps every time. By the end of the exercise set, you may have been wondering ‘isn’t there an easier way to do this?’ The answer is ‘yes.’ In this section, we will derive and use a formula to find the solution of a quadratic equation. We have already seen how to solve a formula for a specific variable ‘in general’ so that we would do the algebraic steps only once and then use the new formula to find the value of the specific variable. Now, we will go through the steps of completing the square in general to solve a quadratic equation for x . It may be helpful to look at one of the examples at the end of the last section where we solved an equation of the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 as you read through the algebraic steps below, so you see them with numbers as well as ‘in general.’
This last equation is the Quadratic Formula.
The solutions to a quadratic equation of the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 , a ≠ 0 a ≠ 0 are given by the formula: To use the Quadratic Formula, we substitute the values of a , b , and c a , b , and c into the expression on the right side of the formula. Then, we do all the math to simplify the expression. The result gives the solution(s) to the quadratic equation. Example 10.28How to solve a quadratic equation using the quadratic formula. Solve 2 x 2 + 9 x − 5 = 0 2 x 2 + 9 x − 5 = 0 by using the Quadratic Formula. Try It 10.55Solve 3 y 2 − 5 y + 2 = 0 3 y 2 − 5 y + 2 = 0 by using the Quadratic Formula. Try It 10.56Solve 4 z 2 + 2 z − 6 = 0 4 z 2 + 2 z − 6 = 0 by using the Quadratic Formula. Solve a quadratic equation using the Quadratic Formula.
If you say the formula as you write it in each problem, you’ll have it memorized in no time. And remember, the Quadratic Formula is an equation. Be sure you start with ‘ x = x = ’. Example 10.29Solve x 2 − 6 x + 5 = 0 x 2 − 6 x + 5 = 0 by using the Quadratic Formula.
Try It 10.57Solve a 2 − 2 a − 15 = 0 a 2 − 2 a − 15 = 0 by using the Quadratic Formula. Try It 10.58Solve b 2 + 10 b + 24 = 0 b 2 + 10 b + 24 = 0 by using the Quadratic Formula. When we solved quadratic equations by using the Square Root Property, we sometimes got answers that had radicals. That can happen, too, when using the Quadratic Formula. If we get a radical as a solution, the final answer must have the radical in its simplified form. Example 10.30Solve 4 y 2 − 5 y − 3 = 0 4 y 2 − 5 y − 3 = 0 by using the Quadratic Formula. We can use the Quadratic Formula to solve for the variable in a quadratic equation, whether or not it is named ‘ x ’.
Try It 10.59Solve 2 p 2 + 8 p + 5 = 0 2 p 2 + 8 p + 5 = 0 by using the Quadratic Formula. Try It 10.60Solve 5 q 2 − 11 q + 3 = 0 5 q 2 − 11 q + 3 = 0 by using the Quadratic Formula. Example 10.31Solve 2 x 2 + 10 x + 11 = 0 2 x 2 + 10 x + 11 = 0 by using the Quadratic Formula.
Try It 10.61Solve 3 m 2 + 12 m + 7 = 0 3 m 2 + 12 m + 7 = 0 by using the Quadratic Formula. Try It 10.62Solve 5 n 2 + 4 n − 4 = 0 5 n 2 + 4 n − 4 = 0 by using the Quadratic Formula. We cannot take the square root of a negative number. So, when we substitute a a , b b , and c c into the Quadratic Formula, if the quantity inside the radical is negative, the quadratic equation has no real solution. We will see this in the next example. Example 10.32Solve 3 p 2 + 2 p + 9 = 0 3 p 2 + 2 p + 9 = 0 by using the Quadratic Formula.
Try It 10.63Solve 4 a 2 − 3 a + 8 = 0 4 a 2 − 3 a + 8 = 0 by using the Quadratic Formula. Try It 10.64Solve 5 b 2 + 2 b + 4 = 0 5 b 2 + 2 b + 4 = 0 by using the Quadratic Formula. The quadratic equations we have solved so far in this section were all written in standard form, a x 2 + b x + c = 0 a x 2 + b x + c = 0 . Sometimes, we will need to do some algebra to get the equation into standard form before we can use the Quadratic Formula. Example 10.33Solve x ( x + 6 ) + 4 = 0 x ( x + 6 ) + 4 = 0 by using the Quadratic Formula.
Try It 10.65Solve x ( x + 2 ) − 5 = 0 x ( x + 2 ) − 5 = 0 by using the Quadratic Formula. Try It 10.66Solve y ( 3 y − 1 ) − 2 = 0 y ( 3 y − 1 ) − 2 = 0 by using the Quadratic Formula. When we solved linear equations, if an equation had too many fractions we ‘cleared the fractions’ by multiplying both sides of the equation by the LCD. This gave us an equivalent equation—without fractions—to solve. We can use the same strategy with quadratic equations. Example 10.34Solve 1 2 u 2 + 2 3 u = 1 3 1 2 u 2 + 2 3 u = 1 3 by using the Quadratic Formula.
Try It 10.67Solve 1 4 c 2 − 1 3 c = 1 12 1 4 c 2 − 1 3 c = 1 12 by using the Quadratic Formula. Try It 10.68Solve 1 9 d 2 − 1 2 d = − 1 2 1 9 d 2 − 1 2 d = − 1 2 by using the Quadratic Formula. Think about the equation ( x − 3 ) 2 = 0 ( x − 3 ) 2 = 0 . We know from the Zero Products Principle that this equation has only one solution: x = 3 x = 3 . We will see in the next example how using the Quadratic Formula to solve an equation with a perfect square also gives just one solution. Example 10.35Solve 4 x 2 − 20 x = −25 4 x 2 − 20 x = −25 by using the Quadratic Formula.
Did you recognize that 4 x 2 − 20 x + 25 4 x 2 − 20 x + 25 is a perfect square? Try It 10.69Solve r 2 + 10 r + 25 = 0 r 2 + 10 r + 25 = 0 by using the Quadratic Formula. Try It 10.70Solve 25 t 2 − 40 t = −16 25 t 2 − 40 t = −16 by using the Quadratic Formula. Use the Discriminant to Predict the Number of Solutions of a Quadratic Equation When we solved the quadratic equations in the previous examples, sometimes we got two solutions, sometimes one solution, sometimes no real solutions. Is there a way to predict the number of solutions to a quadratic equation without actually solving the equation? Yes, the quantity inside the radical of the Quadratic Formula makes it easy for us to determine the number of solutions. This quantity is called the discriminant . DiscriminantIn the Quadratic Formula x = − b ± b 2 − 4 a c 2 a x = − b ± b 2 − 4 a c 2 a , the quantity b 2 − 4 a c b 2 − 4 a c is called the discriminant . Let’s look at the discriminant of the equations in Example 10.28 , Example 10.32 , and Example 10.35 , and the number of solutions to those quadratic equations.
When the discriminant is positive ( x = − b ± + 2 a ) ( x = − b ± + 2 a ) the quadratic equation has two solutions . When the discriminant is zero ( x = − b ± 0 2 a ) ( x = − b ± 0 2 a ) the quadratic equation has one solution . When the discriminant is negative ( x = − b ± − 2 a ) ( x = − b ± − 2 a ) the quadratic equation has no real solutions . Use the discriminant, b 2 − 4 a c b 2 − 4 a c , to determine the number of solutions of a Quadratic Equation.For a quadratic equation of the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 , a ≠ 0 a ≠ 0 ,
Example 10.36Determine the number of solutions to each quadratic equation: ⓐ 2 v 2 − 3 v + 6 = 0 2 v 2 − 3 v + 6 = 0 ⓑ 3 x 2 + 7 x − 9 = 0 3 x 2 + 7 x − 9 = 0 ⓒ 5 n 2 + n + 4 = 0 5 n 2 + n + 4 = 0 ⓓ 9 y 2 − 6 y + 1 = 0 9 y 2 − 6 y + 1 = 0 To determine the number of solutions of each quadratic equation, we will look at its discriminant.
Try It 10.71ⓐ 8 m 2 − 3 m + 6 = 0 8 m 2 − 3 m + 6 = 0 ⓑ 5 z 2 + 6 z − 2 = 0 5 z 2 + 6 z − 2 = 0 ⓒ 9 w 2 + 24 w + 16 = 0 9 w 2 + 24 w + 16 = 0 ⓓ 9 u 2 − 2 u + 4 = 0 9 u 2 − 2 u + 4 = 0 Try It 10.72ⓐ b 2 + 7 b − 13 = 0 b 2 + 7 b − 13 = 0 ⓑ 5 a 2 − 6 a + 10 = 0 5 a 2 − 6 a + 10 = 0 ⓒ 4 r 2 − 20 r + 25 = 0 4 r 2 − 20 r + 25 = 0 ⓓ 7 t 2 − 11 t + 3 = 0 7 t 2 − 11 t + 3 = 0 Identify the Most Appropriate Method to Use to Solve a Quadratic Equation We have used four methods to solve quadratic equations:
You can solve any quadratic equation by using the Quadratic Formula, but that is not always the easiest method to use. Identify the most appropriate method to solve a Quadratic Equation.
What about the method of completing the square? Most people find that method cumbersome and prefer not to use it. We needed to include it in this chapter because we completed the square in general to derive the Quadratic Formula. You will also use the process of completing the square in other areas of algebra. Example 10.37Identify the most appropriate method to use to solve each quadratic equation: ⓐ 5 z 2 = 17 5 z 2 = 17 ⓑ 4 x 2 − 12 x + 9 = 0 4 x 2 − 12 x + 9 = 0 ⓒ 8 u 2 + 6 u = 11 8 u 2 + 6 u = 11 ⓐ 5 z 2 = 17 5 z 2 = 17 Since the equation is in the a x 2 = k a x 2 = k , the most appropriate method is to use the Square Root Property. ⓑ 4 x 2 − 12 x + 9 = 0 4 x 2 − 12 x + 9 = 0 We recognize that the left side of the equation is a perfect square trinomial, and so Factoring will be the most appropriate method. ⓒ 8 u 2 + 6 u = 11 8 u 2 + 6 u = 11 Put the equation in standard form. 8 u 2 + 6 u − 11 = 0 8 u 2 + 6 u − 11 = 0 While our first thought may be to try Factoring, thinking about all the possibilities for trial and error leads us to choose the Quadratic Formula as the most appropriate method Try It 10.73ⓐ x 2 + 6 x + 8 = 0 x 2 + 6 x + 8 = 0 ⓑ ( n − 3 ) 2 = 16 ( n − 3 ) 2 = 16 ⓒ 5 p 2 − 6 p = 9 5 p 2 − 6 p = 9 Try It 10.74ⓐ 8 a 2 + 3 a − 9 = 0 8 a 2 + 3 a − 9 = 0 ⓑ 4 b 2 + 4 b + 1 = 0 4 b 2 + 4 b + 1 = 0 ⓒ 5 c 2 = 125 5 c 2 = 125 Access these online resources for additional instruction and practice with using the Quadratic Formula:
Section 10.3 ExercisesPractice makes perfect. Solve Quadratic Equations Using the Quadratic Formula In the following exercises, solve by using the Quadratic Formula. 4 m 2 + m − 3 = 0 4 m 2 + m − 3 = 0 4 n 2 − 9 n + 5 = 0 4 n 2 − 9 n + 5 = 0 2 p 2 − 7 p + 3 = 0 2 p 2 − 7 p + 3 = 0 3 q 2 + 8 q − 3 = 0 3 q 2 + 8 q − 3 = 0 p 2 + 7 p + 12 = 0 p 2 + 7 p + 12 = 0 q 2 + 3 q − 18 = 0 q 2 + 3 q − 18 = 0 r 2 − 8 r − 33 = 0 r 2 − 8 r − 33 = 0 t 2 + 13 t + 40 = 0 t 2 + 13 t + 40 = 0 3 u 2 + 7 u − 2 = 0 3 u 2 + 7 u − 2 = 0 6 z 2 − 9 z + 1 = 0 6 z 2 − 9 z + 1 = 0 2 a 2 − 6 a + 3 = 0 2 a 2 − 6 a + 3 = 0 5 b 2 + 2 b − 4 = 0 5 b 2 + 2 b − 4 = 0 2 x 2 + 3 x + 9 = 0 2 x 2 + 3 x + 9 = 0 6 y 2 − 5 y + 2 = 0 6 y 2 − 5 y + 2 = 0 v ( v + 5 ) − 10 = 0 v ( v + 5 ) − 10 = 0 3 w ( w − 2 ) − 8 = 0 3 w ( w − 2 ) − 8 = 0 1 3 m 2 + 1 12 m = 1 4 1 3 m 2 + 1 12 m = 1 4 1 3 n 2 + n = − 1 2 1 3 n 2 + n = − 1 2 16 c 2 + 24 c + 9 = 0 16 c 2 + 24 c + 9 = 0 25 d 2 − 60 d + 36 = 0 25 d 2 − 60 d + 36 = 0 5 m 2 + 2 m − 7 = 0 5 m 2 + 2 m − 7 = 0 8 n 2 − 3 n + 3 = 0 8 n 2 − 3 n + 3 = 0 p 2 − 6 p − 27 = 0 p 2 − 6 p − 27 = 0 25 q 2 + 30 q + 9 = 0 25 q 2 + 30 q + 9 = 0 4 r 2 + 3 r − 5 = 0 4 r 2 + 3 r − 5 = 0 3 t ( t − 2 ) = 2 3 t ( t − 2 ) = 2 2 a 2 + 12 a + 5 = 0 2 a 2 + 12 a + 5 = 0 4 d 2 − 7 d + 2 = 0 4 d 2 − 7 d + 2 = 0 3 4 b 2 + 1 2 b = 3 8 3 4 b 2 + 1 2 b = 3 8 1 9 c 2 + 2 3 c = 3 1 9 c 2 + 2 3 c = 3 2 x 2 + 12 x − 3 = 0 2 x 2 + 12 x − 3 = 0 16 y 2 + 8 y + 1 = 0 16 y 2 + 8 y + 1 = 0 In the following exercises, determine the number of solutions to each quadratic equation.
In the following exercises, identify the most appropriate method (Factoring, Square Root, or Quadratic Formula) to use to solve each quadratic equation. Do not solve. ⓐ x 2 − 5 x − 24 = 0 x 2 − 5 x − 24 = 0 ⓑ ( y + 5 ) 2 = 12 ( y + 5 ) 2 = 12 ⓒ 14 m 2 + 3 m = 11 14 m 2 + 3 m = 11 ⓐ ( 8 v + 3 ) 2 = 81 ( 8 v + 3 ) 2 = 81 ⓑ w 2 − 9 w − 22 = 0 w 2 − 9 w − 22 = 0 ⓒ 4 n 2 − 10 = 6 4 n 2 − 10 = 6 ⓐ 6 a 2 + 14 = 20 6 a 2 + 14 = 20 ⓑ ( x − 1 4 ) 2 = 5 16 ( x − 1 4 ) 2 = 5 16 ⓒ y 2 − 2 y = 8 y 2 − 2 y = 8 ⓐ 8 b 2 + 15 b = 4 8 b 2 + 15 b = 4 ⓑ 5 9 v 2 − 2 3 v = 1 5 9 v 2 − 2 3 v = 1 ⓒ ( w + 4 3 ) 2 = 2 9 ( w + 4 3 ) 2 = 2 9 Everyday MathA flare is fired straight up from a ship at sea. Solve the equation 16 ( t 2 − 13 t + 40 ) = 0 16 ( t 2 − 13 t + 40 ) = 0 for t t , the number of seconds it will take for the flare to be at an altitude of 640 feet. An architect is designing a hotel lobby. She wants to have a triangular window looking out to an atrium, with the width of the window 6 feet more than the height. Due to energy restrictions, the area of the window must be 140 square feet. Solve the equation 1 2 h 2 + 3 h = 140 1 2 h 2 + 3 h = 140 for h h , the height of the window. Writing ExercisesSolve the equation x 2 + 10 x = 200 x 2 + 10 x = 200 ⓐ by completing the square ⓑ using the Quadratic Formula ⓒ Which method do you prefer? Why? Solve the equation 12 y 2 + 23 y = 24 12 y 2 + 23 y = 24 ⓐ by completing the square ⓑ using the Quadratic Formula ⓒ Which method do you prefer? Why? ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve? This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Access for free at https://openstax.org/books/elementary-algebra-2e/pages/1-introduction
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What it is, what it does, and how to use it What is the Quadratic Formula?The quadratic formula is:  What does this formula tell us?The quadratic formula calculates the solutions of any quadratic equation. What is a quadratic equation?A quadratic equation is an equation that can be written as ax ² + bx + c where a ≠ 0 . In other words, a quadratic equation must have a squared term as its highest power. Examples of quadratic equations$$ y = 5x^2 + 2x + 5 \\ y = 11x^2 + 22 \\ y = x^2 - 4x +5 \\ y = -x^2 + + 5 $$ Non Examples$$ y = 11x + 22 \\ y = x^3 -x^2 +5x +5 \\ y = 2x^3 -4x^2 \\ y = -x^4 + 5 $$ Ok, but what is a 'solution'?Well a solution can be thought in two ways:
Example of the quadratic formula to solve an equationUse the formula to solve theQuadratic Equation: $$ y = x^2 + 2x + 1 $$ . Just substitute a,b, and c into the general formula: $$ a = 1 \\ b = 2 \\ c = 1 $$ Below is a picture representing the graph of y = x² + 2x + 1 and its solution. Quadratic Formula SongA catchy way to remember the quadratic formula is this song (pop goes the weasel). Practice ProblemsUse the quadratic formula to find the solutions to the following equation: y = x² − 2x + 1 and its solution. Use the quadratic formula to find the solutions to the following equation: y = x² − x − 2 and its solution. In this quadratic equation, y = x² − x − 2 and its solution: Use the quadratic formula to find the solutions to the following equation: y = x² − 1 and its solution. In this quadratic equation, y = x² − 1 and its solution: Calculate the solutions of the the quadratic equation below by using the quadratic formula : y = x² + 2x − 3 and its solution. In this quadratic equation, y = x² + 2x − 3 and its solution: Below is a picture of the graph of the quadratic equation and its two solutions. Use the quadratic formula to find the solutions to the following equation: y = x² + 4x − 5 and its solution. In this quadratic equation, y = x² + 4x − 5 and its solution: Use the quadratic formula to find the solutions to the following equation: y = x² − 4x + 5 and its solution. In this quadratic equation, y = x² − 4x + 5 and its solution: Below is a picture of this quadratic's graph.
Ultimate Math Solver (Free) Free Algebra Solver ... type anything in there!Popular pages @ mathwarehouse.com.  Quadratic Equation SolverWhat do you want to calculate.
Step-By-Step ExampleExample (click to try), choose your method, solve by factoring. Example: 3x^2-2x-1=0 Complete The SquareExample: 3x^2-2x-1=0 (After you click the example, change the Method to 'Solve By Completing the Square'.) Take the Square RootExample: 2x^2=18 Quadratic FormulaExample: 4x^2-2x-1=0 About quadratic equationsNeed more problem types? Try MathPapa Algebra Calculator Clear Quadratic Equation Solver » Quadratic EquationQuadratic equations are second-degree algebraic expressions and are of the form ax 2 + bx + c = 0. The term "quadratic" comes from the Latin word "quadratus" meaning square, which refers to the fact that the variable x is squared in the equation. In other words, a quadratic equation is an “equation of degree 2.” There are many scenarios where a quadratic equation is used. Did you know that when a rocket is launched, its path is described by a quadratic equation? Further, a quadratic equation has numerous applications in physics, engineering, astronomy, etc. Quadratic equations have maximum of two solutions, which can be real or complex numbers. These two solutions (values of x) are also called the roots of the quadratic equations and are designated as (α, β). We shall learn more about the roots of a quadratic equation in the below content.
What is Quadratic Equation?A quadratic equation is an algebraic equation of the second degree in x. The quadratic equation in its standard form is ax 2 + bx + c = 0, where a and b are the coefficients, x is the variable, and c is the constant term. The important condition for an equation to be a quadratic equation is the coefficient of x 2 is a non-zero term (a ≠ 0). For writing a quadratic equation in standard form, the x 2 term is written first, followed by the x term, and finally, the constant term is written.  Further, in real math problems the quadratic equations are presented in different forms: (x - 1)(x + 2) = 0, -x 2 = -3x + 1, 5x(x + 3) = 12x, x 3 = x(x 2 + x - 3). All of these equations need to be transformed into standard form of the quadratic equation before performing further operations. Roots of a Quadratic EquationThe roots of a quadratic equation are the two values of x, which are obtained by solving the quadratic equation. These roots of the quadratic equation are also called the zeros of the equation. For example, the roots of the equation x 2 - 3x - 4 = 0 are x = -1 and x = 4 because each of them satisfies the equation. i.e.,
There are various methods to find the roots of a quadratic equation. The usage of the quadratic formula is one of them. Quadratic FormulaQuadratic formula is the simplest way to find the roots of a quadratic equation . There are certain quadratic equations that cannot be easily factorized, and here we can conveniently use this quadratic formula to find the roots in the quickest possible way. The two roots in the quadratic formula are presented as a single expression. The positive sign and the negative sign can be alternatively used to obtain the two distinct roots of the equation. Quadratic Formula: The roots of a quadratic equation ax 2 + bx + c = 0 are given by x = [-b ± √(b 2 - 4ac)]/2a.  This formula is also known as the Sridharacharya formula . Example: Let us find the roots of the same equation that was mentioned in the earlier section x 2 - 3x - 4 = 0 using the quadratic formula. a = 1, b = -3, and c = -4. x = [-b ± √(b 2 - 4ac)]/2a = [-(-3) ± √((-3) 2 - 4(1)(-4))]/2(1) = [3 ± √25] / 2 = [3 ± 5] / 2 = (3 + 5)/2 or (3 - 5)/2 = 8/2 or -2/2 = 4 or -1 are the roots. Proof of Quadratic FormulaConsider an arbitrary quadratic equation: ax 2 + bx + c = 0, a ≠ 0 To determine the roots of this equation, we proceed as follows: ax 2 + bx = -c ⇒ x 2 + bx/a = -c/a Now, we express the left-hand side as a perfect square , by introducing a new term (b/2a) 2 on both sides:
The left-hand side is now a perfect square: (x + b/2a) 2 = -c/a + b 2 /4a 2 ⇒ (x + b/2a) 2 = (b 2 - 4ac)/4a 2 This is good for us, because now we can take square roots to obtain: x + b/2a = ±√(b 2 - 4ac)/2a x = (-b ± √(b 2 - 4ac))/2a Thus, by completing the squares, we were able to isolate x and obtain the two roots of the equation. Nature of Roots of the Quadratic EquationThe roots of a quadratic equation are usually represented to by the symbols alpha (α), and beta (β). Here we shall learn more about how to find the nature of roots of a quadratic equation without actually finding the roots of the equation. The nature of roots of a quadratic equation can be found without actually finding the roots (α, β) of the equation. This is possible by taking the discriminant value, which is part of the formula to solve the quadratic equation. The value b 2 - 4ac is called the discriminant of a quadratic equation and is designated as 'D'. Based on the discriminant value the nature of the roots of the quadratic equation can be predicted. Discriminant: D = b 2 - 4ac
 Now, check out the formulas to find the sum and the product of the roots of the equation. Sum and Product of Roots of Quadratic EquationThe coefficient of x 2 , x term, and the constant term of the quadratic equation ax 2 + bx + c = 0 are useful in determining the sum and product of the roots of the quadratic equation. The sum and product of the roots of a quadratic equation can be directly calculated from the equation, without actually finding the roots of the quadratic equation. For a quadratic equation ax 2 + bx + c = 0, the sum and product of the roots are as follows.
Writing Quadratic Equations Using RootsThe quadratic equation can also be formed for the given roots of the equation. If α, β, are the roots of the quadratic equation, then the quadratic equation is as follows. x 2 - (α + β)x + αβ = 0 Example: What is the quadratic equation whose roots are 4 and -1? Solution: It is given that α = 4 and β = -1. The corresponding quadratic equation is found by: x 2 - (α + β)x + αβ = 0 x 2 - (α + β)x + αβ = 0 x 2 - (4 - 1)x + (4)(-1) = 0 x 2 - 3x - 4 = 0 Formulas Related to Quadratic EquationsThe following list of important formulas is helpful to solve quadratic equations.
Methods to Solve Quadratic EquationsA quadratic equation can be solved to obtain two values of x or the two roots of the equation. There are four different methods to find the roots of the quadratic equation. The four methods of solving the quadratic equations are as follows.
Method of Completing the Square
Let us look in detail at each of the above methods to understand how to use these methods, their applications, and their uses. Solving Quadratic Equations by FactorizationFactorization of quadratic equation follows a sequence of steps. For a general form of the quadratic equation ax 2 + bx + c = 0, we need to first split the middle term into two terms, such that the product of the terms is equal to the constant term. Further, we can take the common terms from the available term, to finally obtain the required factors as follows:
Here is an example to understand the factorization process.
Thus the two obtained factors of the quadratic equation are (x + 2) and (x + 3). To find its roots, just set each factor to zero and solve for x. i.e., x + 2 = 0 and x + 3 = 0 which gives x = -2 and x = -3. Thus, x = -2 and x = -3 are the roots of x 2 + 5x + 6 = 0. Further, there is another important method of solving a quadratic equation. The method of completing the square for a quadratic equation is also useful to find the roots of the equation. The method of completing the square in a quadratic equation is to algebraically square and simplify, to obtain the required roots of the equation. Consider a quadratic equation ax 2 + bx + c = 0, a ≠ 0. To determine the roots of this equation, we simplify it as follows:
Now, we express the left-hand side as a perfect square, by introducing a new term (b/2a) 2 on both sides:
Here the '+' sign gives one root and the '-' sign gives another root of the quadratic equation. Generally, this detailed method is avoided, and only the quadratic formula is used to obtain the required roots. Graphing a Quadratic Equation The point(s) where the graph cuts the horizontal x-axis (typically the x-intercepts ) is the solution of the quadratic equation. These points can also be algebraically obtained by equalizing the y value to 0 in the function y = ax 2 + bx + c and solving for x. Quadratic Equations Having Common RootsConsider two quadratic equations having common roots a 1 x 2 + b 1 x + c 1 = 0, and a 2 x 2 + b 2 x + c 2 = 0. Let us solve these two equations to find the conditions for which these equations have a common root. The two equations are solved for x 2 and x respectively. (x 2 )(b 1 c 2 - b 2 c 1 ) = (-x)/(a 1 c 2 - a 2 c 1 ) = 1/(a 1 b 2 - a 2 b 1 ) x 2 = (b 1 c 2 - b 2 c 1 ) / (a 1 b 2 - a 2 b 1 ) x = (a 2 c 1 - a 1 c 2 ) / (a 1 b 2 - a 2 b 1 ) Hence, by simplifying the above two expressions we have the following condition for the two equations having the common root. (a 1 b 2 - a 2 b 1 ) (b 1 c 2 - b 2 c 1 ) = (a 2 c 1 - a 1 c 2 ) 2 Maximum and Minimum Value of Quadratic Expression The maximum and minimum values of the quadratic expressions are of further help to find the range of the quadratic expression: The range of the quadratic expressions also depends on the value of a. For positive values of a( a > 0), the range is [ F(-b/2a), ∞), and for negative values of a ( a < 0), the range is (-∞, F(-b/2a)].
Note that the domain of a quadratic function is the set of all real numbers, i.e., (-∞, ∞). Tips and Tricks on Quadratic Equation: Some of the below-given tips and tricks on quadratic equations are helpful to more easily solve quadratic equations.
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Cuemath is one of the world's leading math learning platforms that offers LIVE 1-to-1 online math classes for grades K-12 . Our mission is to transform the way children learn math, to help them excel in school and competitive exams. Our expert tutors conduct 2 or more live classes per week, at a pace that matches the child's learning needs. Quadratic Equations ExamplesExample 1: James is a fitness enthusiast and goes for a jog every morning. The park where he jogs is rectangular in shape and measures 12m × 8m. An environmentalist group plans to revamp the park and decides to build a pathway surrounding the park. This would increase the total area to 140 sq m. What will be the width of the pathway? Let’s denote the width of the pathway by x. Then, the length and breadth of the outer rectangle is (12+2x) m and (8+2x) m. Given that, area = 140 (12 + 2x)(8 + 2x) = 140 2(6 + x) 2(4 + x) = 140 (6 + x)(4 + x) = 35 24 + 6x + 4x + x 2 = 35 x 2 + 10x -11 = 0 x 2 + 11x - x - 11 = 0 x(x + 11) - 1(x + 11) = 0 (x + 11)(x - 1) = 0 (x + 11) =0 and (x - 1) = 0 x = -11 and x = 1 Since length can’t be negative, we take x = 1. Answer: Therefore the width of the pathway is 1 m. Example 2: Rita throws a ball upwards from a platform that is 20m above the ground. The height of the ball from the ground at a time 't', is denoted by 'h'. Suppose h = -4t 2 + 16t + 20. Find the maximum height attained by the ball. We can rearrange the terms of the quadratic equation h = -4t 2 + 16t + 20 in such a way that it is easy to find the maximum value of this equation. = -4t 2 + 16t + 20 = -4(t 2 - 4t - 5) = -4((t - 2) 2 - 9) = -4(t - 2) 2 + 36 We should keep the value of (t - 2) 2 minimum in order to find the maximum value of h. So, the minimum value (t - 2) 2 can take is 0. Answer: Therefore the maximum height attained is 36m. Example 3: Find the quadratic equation having the roots 5 and 8 respectively. The quadratic equation having the roots α, β, is x 2 - (α + β)x + αβ = 0. Given α = 5, and β = 8. Therefore the quadratic equation is: x 2 - (5 + 8)x + 5×8 = 0 x 2 - 13x + 40 = 0 Answer: Hence the required quadratic equation is x 2 - 13x + 40 = 0 Example 4: The quad equation 2x 2 + 9x + 7 = 0 has roots α, β. Find the quadratic equation having the roots 1/α, and 1/β. The quadratic equation having roots that are reciprocal to the roots of the equation ax 2 + bx + c = 0, is cx 2 + bx + a = 0. The given quadratic equation is 2x 2 + 9x + 7 = 0. Hence the required equation having reciprocal roots is 7x 2 + 9x + 2 = 0. Method 2: From the given equation, α + β = -9/2 and α β = 2/7. The new equation should have its roots to be 1/α and 1/β. Their sum = 1/α + 1/β = (α + β) / α β = -9/7 Their product = 1/α β = 2/7 Thus, the required equation is, x 2 - (1/α + 1/β)x + 1/α β = 0 x 2 - (-9/7)x + 2/7 = 0 Multiplying both sides by 7, 7x 2 + 9x + 2 = 0 Answer: Therefore the equation is 7x 2 + 9x + 2 = 0. go to slide go to slide go to slide go to slide  Book a Free Trial Class Practice Questions on Quadratic Equationgo to slide go to slide FAQs on Quadratic EquationWhat is the definition of a quadratic equation. A quadratic equation in math is a second-degree equation of the form ax 2 + bx + c = 0. Here a and b are the coefficients, c is the constant term, and x is the variable. Since the variable x is of the second degree, there are two roots or answers for this quadratic equation. The roots of the quadratic equation can be found by either solving by factorizing or through the use of the quadratic formula. What is the Quadratic Formula?The quadratic equation formula to solve the equation ax 2 + bx + c = 0 is x = [-b ± √(b 2 - 4ac)]/2a. Here we obtain the two values of x, by applying the plus and minus symbols in this formula. Hence the two possible values of x are [-b + √(b 2 - 4ac)]/2a, and [-b - √(b 2 - 4ac)]/2a. How do You Solve a Quadratic Equation?There are several methods to solve quadratic equations, but the most common ones are factoring, using the quadratic formula, and completing the square.
What is Determinant in Quadratic Formula?The value b 2 - 4ac is called the discriminant and is designated as D. The discriminant is part of the quadratic formula. The discriminants help us to find the nature of the roots of the quadratic equation, without actually finding the roots of the quadratic equation. What are Some Real-Life Applications of Quadratic Equations?Quadratic equations are used to find the zeroes of the parabola and its axis of symmetry . There are many real-world applications of quadratic equations.
How are Quadratic Equations Different From Linear Equations?A linear degree is an equation of a single degree and one variable, and a quadratic equation is an equation in two degrees and a single variable. A linear equation is of the the form ax + b = 0 and a quadratic equation is of the form ax 2 + bx + c = 0. A linear equation has a single root and a quadratic equation has two roots or two answers. Also, a quadratic equation is a product of two linear equations. What Are the 4 Ways To Solve A Quadratic Equation?The four ways of solving a quadratic equation are as follows.
How to Solve a Quadratic Equation by Completing the Square?The quadratic equation is solved by the method of completing the square and it uses the formula (a + b)^2 = a 2 + 2ab + b 2 (or) (a - b)^2 = a 2 - 2ab + b 2 . How to Find the Value of the Discriminant?The value of the discriminant in a quadratic equation can be found from the variables and constant terms of the standard form of the quadratic equation ax 2 + bx + c = 0. The value of the discriminant is D = b 2 - 4ac, and it helps to predict the nature of roots of the quadratic equation, without actually finding the roots of the equation. How Do You Solve Quadratic Equations With Graphing?The quadratic equation can be solved similarly to a linear equal by graphing. Let us take the quadratic equation ax 2 + bx + c = 0 as y = ax 2 + bx + c . Here we take the set of values of x and y and plot the graph. The two points where this graph meets the x-axis, are the solutions of this quadratic equation. How Important Is the Discriminant of a Quadratic Equation?The discriminant is very much needed to easily find the nature of the roots of the quadratic equation. Without the discriminant, finding the nature of the roots of the equation is a long process, as we first need to solve the equation to find both the roots. Hence the discriminant is an important and needed quantity, which helps to easily find the nature of the roots of the quadratic equation. Where Can I Find Quadratic Equation Solver?To get the quadratic equation solver, click here . Here, we can enter the values of a, b, and c for the quadratic equation ax 2 + bx + c = 0, then it will give you the roots along with a step-by-step procedure. What is the Use of Discriminants in Quadratic Formula?The discriminant (D = b 2 - 4ac) is useful to predict the nature of the roots of the quadratic equation. For D > 0, the roots are real and distinct, for D = 0 the roots are real and equal, and for D < 0, the roots do not exist or the roots are imaginary complex numbers . With the help of this discriminant and with the least calculations, we can find the nature of the roots of the quadratic equation. How do you Solve a Quadratic Equation without Using the Quadratic Formula?There are two alternative methods to the quadratic formula. One method is to solve the quadratic equation through factorization, and another method is by completing the squares. In total there are three methods to find the roots of a quadratic equation. How to Derive Quadratic Formula?The algebra formula (a + b) 2 = a 2 + 2ab + b 2 is used to solve the quadratic equation and derive the quadratic formula. This algebraic formula is used to manipulate the quadratic equation and derive the quadratic formula to find the roots of the equation. Quadratic Formula CalculatorEnter the equation you want to solve using the quadratic formula. The Quadratic Formula Calculator finds solutions to quadratic equations with real coefficients. For equations with real solutions, you can use the graphing tool to visualize the solutions. Quadratic Formula : x = − b ± b 2 − 4 a c 2 a Click the blue arrow to submit. Choose "Solve Using the Quadratic Formula" from the topic selector and click to see the result in our Algebra Calculator ! Solve Using the Quadratic Formula Apply the Quadratic Formula Popular ProblemsSolve Using the Quadratic Formula x 2 + 5 x + 6 = 0 Solve Using the Quadratic Formula x 2 - 9 = 0 Solve Using the Quadratic Formula 5 x 2 - 7 x - 3 = 0 Apply the Quadratic Formula x 2 - 14 x + 49 Apply the Quadratic Formula x 2 - 18 x - 4 Please ensure that your password is at least 8 characters and contains each of the following:
 Calculator Soup ® Online Calculators Quadratic Formula CalculatorCalculator use. This online calculator is a quadratic equation solver that will solve a second-order polynomial equation such as ax 2 + bx + c = 0 for x, where a ≠ 0, using the quadratic formula . The calculator solution will show work using the quadratic formula to solve the entered equation for real and complex roots. Calculator determines whether the discriminant \( (b^2 - 4ac) \) is less than, greater than or equal to 0. When \( b^2 - 4ac = 0 \) there is one real root. When \( b^2 - 4ac > 0 \) there are two real roots. When \( b^2 - 4ac < 0 \) there are two complex roots. Quadratic Formula:The quadratic formula is used to solve quadratic equations where a ≠ 0 (polynomials with an order of 2) Examples using the quadratic formulaExample 1: Find the Solution for \( x^2 + -8x + 5 = 0 \), where a = 1, b = -8 and c = 5, using the Quadratic Formula. The discriminant \( b^2 - 4ac > 0 \) so, there are two real roots. Simplify the Radical: Simplify fractions and/or signs: which becomes Example 2: Find the Solution for \( 5x^2 + 20x + 32 = 0 \), where a = 5, b = 20 and c = 32, using the Quadratic Formula. The discriminant \( b^2 - 4ac < 0 \) so, there are two complex roots. calculator updated to include full solution for real and complex roots Cite this content, page or calculator as: Furey, Edward " Quadratic Formula Calculator " at https://www.calculatorsoup.com/calculators/algebra/quadratic-formula-calculator.php from CalculatorSoup, https://www.calculatorsoup.com - Online Calculators Last updated: August 17, 2023 Pardon Our InterruptionAs you were browsing something about your browser made us think you were a bot. There are a few reasons this might happen:
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Number LineThe Algebra Calculator is a versatile online tool designed to simplify algebraic problem-solving for users of all levels. Here's how to make the most of it:
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Solve each equation with the quadratic formula. 1) v2 + 2v − 8 = 0 2) k2 + 5k − 6 = 0 ... Infinite Algebra 2 Name_____ Using the Quadratic Formula Date_____ Period____ Solve each equation with the quadratic formula. 1) v2 + 2v − 8 = 0 {2, −4} 2) k2 + 5k − 6 = 0 {1, −6} 3) 2v2 − 5v + 3 = 0 {3 ...
0 points each Assignment: Solve the problems below. Write the problem, your work, and the solution in the text box below to submit your work. Be sure to show all of your work. Here is a link explaining how to show your work. We suggest saving your work in a word processor. Solve each problem below showing the steps as indicated in the lesson. 1.
Vertex form. Graphing quadratic inequalities. Factoring quadratic expressions. Solving quadratic equations w/ square roots. Solving quadratic equations by factoring. Completing the square. Solving equations by completing the square. Solving equations with the quadratic formula. The discriminant.
This algebra 2 video tutorial focuses on solving quadratic equations using the quadratic formula. It also explains how to solve quadratic equations by facto...
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Create your own worksheets like this one with Infinite Algebra 2. Free trial available at KutaSoftware.com. ©Q D2x0o1S2P iKSuGtRa6 4S1oGf1twwuamrUei 0LjLoCM.W T PAMlcl4 drhisg2hatEsB XrqeQsger KvqeidM.2 v 5M1awdPeZ uwjirtbhi QIxnDftiFn4iOteeE qAwlXg1ezbor9aP u2B.w.
Section 6.4 Quadratic Formula. A2.5.6 Describe characteristics of quadratic functions and use them to solve real-world problems. Need a tutor? Click this link and get your first session free! ... Download File. Practice Solutions. a2_6.4_practice_solutions.pdf: File Size: 212 kb: File Type: pdf: Download File. Corrective Assignment. a2_6.4_ca ...
Rewrite the quadratic equation in the standard form. Convert the quadratic equation into the standard form. This will get rid of the denominators thereby giving us integer values for. Stay informed about the latest lessons as they become available on our website. Below are ten (10) practice problems regarding the quadratic formula.
Algebra 2 Lessons and Practice is a free site for students (and teachers) studying a second year of high school algebra. ... The quadratic formula, however, may be used to solve ANY quadratic equation (even the ones that can be factored). This is a formula that you want to know and remember! ... Solution using Quadratic Formula: a = 2; b = -10 ...
Intermediate Algebra 2e 9.3 Solve Quadratic Equations Using the Quadratic Formula. ... We will see in the next example how using the Quadratic Formula to solve an equation whose standard form is a perfect square trinomial equal to 0 gives just one solution. Notice that once the radicand is simplified it becomes 0 , which leads to only one ...
Level 2 exercises, Solving Quadratic Equations With the Quadratic Formula. Level 2 - Exercises that is reviewed to match approximately C-B level. The function p(x)=x^2-2x-15 has been graphed below along with a straight line. The straight line passes through the vertex and one of the x-intercepts of
Enjoy these free sheets. Each one has model problems worked out step by step, practice problems, as well as challenge questions at the sheets end. Plus each one comes with an answer key. Solve Quadratic Equations by Factoring. Solve Quadratic Equations by Completing the Square. Quadratic Formula Worksheets.
The solutions to a quadratic equation of the form ax2 + bx + c = 0 a x 2 + b x + c = 0, a ≠ 0 a ≠ 0 are given by the formula: To use the Quadratic Formula, we substitute the values of a, b, andc a, b, and c into the expression on the right side of the formula. Then, we do all the math to simplify the expression.
Example of the quadratic formula to solve an equation. Use the formula to solve theQuadratic Equation: y = x2 + 2x + 1 y = x 2 + 2 x + 1. Just substitute a,b, and c into the general formula: a = 1 b = 2 c = 1 a = 1 b = 2 c = 1. Below is a picture representing the graph of y = x² + 2x + 1 and its solution.
To solve a quadratic equation, use the quadratic formula: x = (-b ± √ (b^2 - 4ac)) / (2a). There can be 0, 1 or 2 solutions to a quadratic equation. If the discriminant is positive there are two solutions, if negative there is no solution, if equlas 0 there is 1 solution. In math, a quadratic equation is a second-order polynomial equation in ...
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
A quadratic equation is an algebraic equation of the second degree in x. The quadratic equation in its standard form is ax 2 + bx + c = 0, where a and b are the coefficients, x is the variable, and c is the constant term. The important condition for an equation to be a quadratic equation is the coefficient of x 2 is a non-zero term (a ≠ 0). For writing a quadratic equation in standard form ...
Step 1: Enter the equation you want to solve using the quadratic formula. The Quadratic Formula Calculator finds solutions to quadratic equations with real coefficients. For equations with real solutions, you can use the graphing tool to visualize the solutions. Quadratic Formula: x = −b±√b2 −4ac 2a x = − b ± b 2 − 4 a c 2 a.
Calculator Use. This online calculator is a quadratic equation solver that will solve a second-order polynomial equation such as ax 2 + bx + c = 0 for x, where a ≠ 0, using the quadratic formula. The calculator solution will show work using the quadratic formula to solve the entered equation for real and complex roots.
Algebra 2 Assignment 3.2 Welcome to Assignment 3.2. In this set of problems, you will solve equations, simplify expressions, and explore various algebraic functions. Remember to justify each step of your solution, explaining your reasoning thoroughly. Providing a clear and well-reasoned justification is just as important as finding the correct answer.
The Algebra Calculator is a versatile online tool designed to simplify algebraic problem-solving for users of all levels. Here's how to make the most of it: Begin by typing your algebraic expression into the above input field, or scanning the problem with your camera. After entering the equation, click the 'Go' button to generate instant solutions.