rational inequalities assignment quizlet edgenuity

Module 1: Equations and Inequalities

▪ solving rational inequalities, learning outcomes.

• Solve rational inequalities using boundary value method.

Solving Rational Inequalities using Boundary Value Method

Any inequality that can be put into one of the following forms

[latex]f(x)>0, f(x) \geq 0, f(x)<0, or f(x) \leq 0[/latex], where [latex]f[/latex] is a rational function

is called rational inequality .

How To: Solve Rational Inequalities

• Find restriction(s) of the given rational expression(s) in the inequality. The restriction(s) is(are) the boundary point(s) .
• Rewrite the given rational inequality as an equation by replacing the inequality symbol with the equal sign.
• Solve the rational equation. The real solution(s) of the equation is(are) the boundary point(s) .
• Plot the boundary point(s) from Step 1 & 3 on a number line. [latex] \Rightarrow [/latex] Use an open circle ALL restrictions. [latex] \Rightarrow [/latex] Use an open circle when the given inequality has [latex]<[/latex] or [latex]>[/latex] [latex]\Rightarrow[/latex] Use a closed circle when the given inequality has [latex]\leq[/latex] or [latex]\geq[/latex].
• Choose one number, which is called a test value , from each interval and test the intervals by evaluating the given inequality at that number. [latex]\Rightarrow[/latex] If the inequality is TRUE, then the interval is a solution of the inequality. [latex]\Rightarrow[/latex] If the inequality is FALSE, then the interval is not a solution of the inequality.
• Write the solution set (usually in interval notation), selecting the interval(s) from Step 5.

Example: Solving Rational Inequality using Boundary Value Method

Solve the rational inequality using boundary value method. Graph the solution set and write the solution in interval notation.

[latex]\frac{2x-1}{x+2} > 1[/latex]

First, find restriction(s):

[latex]x+2 \ne 0[/latex]

[latex]x \ne -2[/latex]

Second, let’s rewrite the inequality as an equation and then solve it:

[latex]\frac{2x-1}{x+2} = 1[/latex]

[latex]\frac{2x-1}{x+2} = \frac{1}{1}[/latex]

[latex]2x-1=x+2[/latex]

[latex]x=3[/latex]

Third, plot the restriction and the solution on the number line. All restrictions should be an  open circle. Since the inequality has “[latex]>[/latex]” (greater than), which doesn’t have “[latex]=[/latex]” (equal sign), the solution should be an  open circle as well:

Note that now there are three intervals on the number line: [latex](-\infty, -2), (-2, 3),[/latex] and [latex](3, \infty)[/latex].

Fourth, choose your choice of test value from each interval:

I chose [latex]-3, 0,[/latex] and [latex]4[/latex] from those three intervals.

fifth, test each interval using the test value:

When [latex]x=-3[/latex], [latex]\frac{2(-3)-1}{(-3)+2} > 1 \rightarrow \frac{-7}{-1}>1 \rightarrow 7>1[/latex]  (True)   [latex]\therefore[/latex] solution

When [latex]x=-3[/latex], [latex]\frac{2(0)-1}{(0)+2} > 1 \rightarrow \frac{-1}{2}>1 \rightarrow -\frac{1}{2}>1[/latex]  (False)   [latex]\therefore[/latex] not a solution

When [latex]x=4[/latex], [latex]\frac{2(4)-1}{(4)+2} > 1 \rightarrow \frac{7}{6}>1[/latex]  (True)   [latex]\therefore[/latex] solution

Actually, we can check the results of test values from the graph of [latex]f(x)=\frac{2x-1}{x+2}-1[/latex] below:

Therefore, the solution of the rational inequality [latex]\frac{2x-1}{x+2} > 1[/latex] is [latex](-\infty, -2) \cup (3, \infty)[/latex].

[latex]\frac{5x}{x-3} \leq 2[/latex]

[latex]x-3 \ne 0[/latex]

[latex]x \ne 3[/latex]

[latex]\frac{5x}{x-3}=2[/latex]

[latex]\frac{5x}{x-3} = \frac{2}{1}[/latex]

[latex]5x=2(x-3)[/latex]

[latex]5x=2x-6[/latex]

[latex]3x=-6[/latex]

[latex]x=-2[/latex]

Third, plot the restriction and the solution on the number line. All restrictions should be an  open circle. Since the inequality has “[latex]\leq[/latex]” (less than or equal to), which has “[latex]=[/latex]” (equal sign), the solution should be a  closed circle:

Note that now there are three intervals on the number line: [latex](-\infty, -2], [-2, 3),[/latex] and [latex](3, \infty)[/latex].

When [latex]x=-3[/latex], [latex]\frac{5(-3)}{(-3)-3} \leq 2 \rightarrow \frac{-15}{-6} \leq 2 \rightarrow 2.5 \leq 2[/latex]  (False)   [latex]\therefore[/latex] not a solution

When [latex]x=0[/latex], [latex]\frac{5(0)}{(0)-3} \leq 2 \rightarrow \frac{0}{-3} \leq 2 \rightarrow 0 \leq 2[/latex]  (True)   [latex]\therefore[/latex] solution

When [latex]x=4[/latex], [latex]\frac{5(4)}{(4)-3} \leq 2 \rightarrow \frac{20}{1} \leq 2 \rightarrow 20 \leq 2[/latex]  (False)   [latex]\therefore[/latex] not a solution

Actually, we can check the results of test values from the graph of [latex]f(x)=\frac{5x}{x-3}-2[/latex] below:

Therefore, the solution of the rational inequality [latex]\frac{5x}{x-3} \leq 2[/latex] is [latex][-2, 3)[/latex].

[latex]\frac{4x-5}{x+1} \geq 0[/latex]

[latex]x+1 \ne 0[/latex]

[latex]x \ne -1[/latex]

[latex]\frac{4x-5}{x+1}=0[/latex]

[latex]\frac{4x-5}{x+1} = \frac{0}{1}[/latex]

[latex]4x-5=0[/latex]

[latex]4x=5[/latex]

[latex]x=\frac{5}{4}[/latex]

Third, plot the restriction and the solution on the number line. All restrictions should be an  open circle. Since the inequality has “[latex]\geq[/latex]” (greater than or equal to), which has “[latex]=[/latex]” (equal sign), the solution should be a  closed circle:

Note that now there are three intervals on the number line: [latex](-\infty, -1), (-1, \frac{5}{4}],[/latex] and [latex][\frac{5}{4}, \infty)[/latex].

I chose [latex]-2, 0,[/latex] and [latex]2[/latex] from those three intervals.

When [latex]x=-2[/latex], [latex]\frac{4(-2)-5}{(-2)+1} \geq 0 \rightarrow \frac{-13}{-1} \geq 0 \rightarrow 13 \geq 0[/latex]  (True)   [latex]\therefore[/latex] solution

When [latex]x=0[/latex], [latex]\frac{4(0)-5}{(0)+1} \geq 0 \rightarrow \frac{-5}{1} \geq 0 \rightarrow -5 \geq 0[/latex]  (False)   [latex]\therefore[/latex] not a solution

When [latex]x=2[/latex], [latex]\frac{4(2)-5}{(2)+1} \geq 0 \rightarrow \frac{3}{3} \geq 0 \rightarrow 1 \geq 0[/latex]  (True)   [latex]\therefore[/latex] solution

Actually, we can check the results of test values from the graph of [latex]f(x)=\frac{4x-5}{x+1}[/latex] below:

Therefore, the solution of the rational inequality [latex]\frac{4x-5}{x+1} \geq 0[/latex] is [latex](-\infty, -1) \cup [\frac{5}{4}, \infty)[/latex].

• Solving Rational Inequalities. Authored by : Michelle Eunhee Chung. Provided by : Georgia State University. License : CC BY: Attribution

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5.7: Solve Quadratic Inequalities

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Learning Objectives

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

• Solve quadratic inequalities graphically
• Solve quadratic inequalities algebraically

Before you get started, take this readiness quiz.

• Solve: \(2x−3=0\). If you missed this problem, review Example 2.2.
• Solve: \(2y^{2}+y=15\). If you missed this problem, review Example 6.45.
• Solve \(\frac{1}{x^{2}+2 x-8}>0\) If you missed this problem, review Example 7.56.

We have learned how to solve linear inequalities and rational inequalities previously. Some of the techniques we used to solve them were the same and some were different. We will now learn to solve inequalities that have a quadratic expression. We will use some of the techniques from solving linear and rational inequalities as well as quadratic equations. We will solve quadratic inequalities two ways—both graphically and algebraically.

Solve Quadratic Inequalities Graphically

A quadratic equation is in standard form when written as \(ax^{2}+bx+c=0\). If we replace the equal sign with an inequality sign, we have a quadratic inequality in standard form.

Definition \(\PageIndex{1}\): Quadratic Inequality

A quadratic inequality is an inequality that contains a quadratic expression. The standard form of a quadratic inequality is written:

\(\begin{array}{ll}{a x^{2}+b x+c<0} & {a x^{2}+b x+c \leq 0} \\ {a x^{2}+b x+c>0} & {a x^{2}+b x+c \geq 0}\end{array}\)

The graph of a quadratic function \(f(x)=a x^{2}+b x+c=0\) is a parabola. When we ask when is \(a x^{2}+b x+c<0\), we are asking when is \(f(x)<0\). We want to know when the parabola is below the \(x\)-axis.

When we ask when is \(a x^{2}+b x+c>0\), we are asking when is \(f(x)>0\). We want to know when the parabola is above the \(y\)-axis.

Example \(\PageIndex{1}\): How to Solve a Quadratic Inequality Graphically

Solve \(x^{2}−6x+8<0\) graphically. Write the solution in interval notation.

Step 1 : Write the quadratic inequality in standard form.

The inequality is in standard form.

\(x^{2}-6 x+8<0\)

Step 2 : Graph the function \(f(x)=a x^{2}+b x+c\) using properties or transformations.

We will graph using the properties.

\(f(x)=x^{2}-6 x+8\)

Look at \(a\) in the equation.

\(\color{red}{a=1, b=-6, c=8}\)

Since \(a\) is positive, the parabola opens upward.

The parabola opens upward.

The axis of symmetry is the line \(x=-\frac{b}{2 a}\).

Axis of Symmetry

\(x=-\frac{b}{2 a}\)

\(\begin{array}{l}{x=-\frac{(-6)}{2 \cdot 1}} \\ {x=3}\end{array}\)

The axis of symmetry is the line \(x=3\).

The vertex is on the axis of symmetry. Substitute \(x=3\) into the function.

\(\begin{array}{l}{f(x)=x^{2}-6 x+8} \\ {f(3)=(\color{red}{3}\color{black}{)}^{2}-6(\color{red}{3}\color{black}{)}+8} \\ {f(3)=-1}\end{array}\)

The vertex is \((3,-1)\).

We find \(f(0)\)

\(y\)-intercept

\(\begin{array}{l}{f(x)=x^{2}-6 x+8} \\ {f(0)=(\color{red}{0}\color{black}{)}^{2}-6(\color{red}{0}\color{black}{)}+8} \\ {f(0)=8}\end{array}\)

The \(y\)-intercept is \((0,8)\).

We use the axis of symmetry to find a point symmetric to the \(y\)-intercept. The \(y\)-intercept is \(3\) units left of the axis of symmetry, \(x=3\). A point \(3\) units to the right of the axis of symmetry has \(x=6\).

Point symmetric to \(y\)-intercept

The point is \((6,8)\).

We solve \(f(x)=0\).

\(x\)-intercepts

We can solve this quadratic equation by factoring.

\(\begin{aligned} f(x) &=x^{2}-6 x+8 \\ \color{red}{0} &\color{black}{=}x^{2}-6 x+8 \\ \color{red}{0} &\color{black}{=}(x-2)(x-4) \\ x &=2 \text { or } x=4 \end{aligned}\)

The \(x\)-intercepts are \((2,0)\) and \((4,0)\).

We graph the vertex, intercepts, and the point symmetric to the \(y\)-intercept. We connect these \(5\) points to sketch the parabola.

Step 3 : Determine the solution from the graph.

The inequality asks for the values of \(x\) which make the function less than \(0\). Which values of \(x\) make the parabola below the \(x\)-axis.

We do not include the values \(2\), \(4\) as the inequality is less than only.

The solution, in interval notation, is \((2,4)\).

Exercise \(\PageIndex{1}\)

• Solve \(x^{2}+2 x-8<0\) graphically
• Write the solution in interval notation

Exercise \(\PageIndex{2}\)

• Solve \(x^{2}-8 x+12 \geq 0\) graphically

• \((-\infty, 2] \cup[6, \infty)\)

We list the steps to take to solve a quadratic inequality graphically.

Solve a Quadratic Inequality Graphically

• Write the quadratic inequality in standard form.
• Graph the function \(f(x)=ax^{2}+bx+c\).
• Determine the solution from the graph.

In the last example, the parabola opened upward and in the next example, it opens downward. In both cases, we are looking for the part of the parabola that is below the \(x\)-axis but note how the position of the parabola affects the solution.

Example \(\PageIndex{2}\)

Solve \(-x^{2}-8 x-12 \leq 0\) graphically. Write the solution in interval notation.

Exercise \(\PageIndex{3}\)

• Solve \(-x^{2}-6 x-5>0\) graphically

• \((-5,-1)\)

Exercise \(\PageIndex{4}\)

• Solve \(−x^{2}+10x−16≤0\) graphically

• \((-\infty, 2] \cup[8, \infty)\)

Solve Quadratic Inequalities Algebraically

The algebraic method we will use is very similar to the method we used to solve rational inequalities. We will find the critical points for the inequality, which will be the solutions to the related quadratic equation. Remember a polynomial expression can change signs only where the expression is zero.

We will use the critical points to divide the number line into intervals and then determine whether the quadratic expression will be positive or negative in the interval. We then determine the solution for the inequality.

Example \(\PageIndex{3}\): How to Solve Quadratic Inequalities Algebraically

Solve \(x^{2}-x-12 \geq 0\) algebraically. Write the solution in interval notation.

Exercise \(\PageIndex{5}\)

Solve \(x^{2}+2x−8≥0\) algebraically. Write the solution in interval notation.

\((-\infty,-4] \cup[2, \infty)\)

Exercise \(\PageIndex{6}\)

Solve \(x^{2}−2x−15≤0\) algebraically. Write the solution in interval notation.

In this example, since the expression \(x^{2}−x−12\) factors nicely, we can also find the sign in each interval much like we did when we solved rational inequalities. We find the sign of each of the factors, and then the sign of the product. Our number line would like this:

The result is the same as we found using the other method.

We summarize the steps here.

Solve a Quadratic Inequality Algebraically

• Determine the critical points—the solutions to the related quadratic equation.
• Use the critical points to divide the number line into intervals.
• Above the number line show the sign of each quadratic expression using test points from each interval substituted into the original inequality.
• Determine the intervals where the inequality is correct. Write the solution in interval notation.

Example \(\PageIndex{4}\)

Solve \(x^{2}+6x−7≥0\) algebraically. Write the solution in interval notation.

Exercise \(\PageIndex{7}\)

Solve \(−x^{2}+2x+1≥0\) algebraically. Write the solution in interval notation.

\([-1-\sqrt{2},-1+\sqrt{2}]\)

Exercise \(\PageIndex{8}\)

Solve \(−x^{2}+8x−14<0\) algebraically. Write the solution in interval notation.

\((-\infty, 4-\sqrt{2}) \cup(4+\sqrt{2}, \infty)\)

The solutions of the quadratic inequalities in each of the previous examples, were either an interval or the union of two intervals. This resulted from the fact that, in each case we found two solutions to the corresponding quadratic equation \(ax^{2}+bx+c=0\). These two solutions then gave us either the two \(x\) - intercepts for the graph or the two critical points to divide the number line into intervals.

This correlates to our previous discussion of the number and type of solutions to a quadratic equation using the discriminant.

For a quadratic equation of the form \(ax^{2}+bc+c=0, a≠0\).

The last row of the table shows us when the parabolas never intersect the \(x\)-axis. Using the Quadratic Formula to solve the quadratic equation, the radicand is a negative. We get two complex solutions.

In the next example, the quadratic inequality solutions will result from the solution of the quadratic equation being complex.

Example \(\PageIndex{5}\)

Solve, writing any solution in interval notation:

• \(x^{2}-3 x+4>0\)

\(x^{2}-3 x+4 \leq 0\)

We are to find the solution to \(x^{2}−3x+4>0\). Since for all values of \(x\) the graph is above the \(x\)-axis, all values of \(x\) make the inequality true. In interval notation we write \((−∞,∞)\).

b. Write the quadratic inequality in standard form.

Determine the critical points by solving the related quadratic equation.

\(x^{2}-3 x+4=0\)

Since the corresponding quadratic equation is the same as in part (a), the parabola will be the same. The parabola opens upward and is completely above the \(x\)-axis—no part of it is below the \(x\)-axis.

We are to find the solution to \(x^{2}−3x+4≤0\). Since for all values of \(x\) the graph is never below the \(x\)-axis, no values of \(x\) make the inequality true. There is no solution to the inequality.

Exercise \(\PageIndex{9}\)

Solve and write any solution in interval notation:

• \(-x^{2}+2 x-4 \leq 0\)
• \(-x^{2}+2 x-4 \geq 0\)
• \((-\infty, \infty)\)
• no solution

Exercise \(\PageIndex{10}\)

• \(x^{2}+3 x+3<0\)
• \(x^{2}+3 x+3>0\)

Key Concepts

• Graph the function \(f(x)=ax^{2}+bx+c\) using properties or transformations.
• Determine the critical points -- the solutions to the related quadratic equation.