Section1.7Quadratic and Rational Inequalities (EQ7)
Objectives
Solve quadratic inequalities and express the solution graphically and with interval notation. Solve rational inequalities and express the solution graphically and using interval notation.
Subsection1.7.1Activities
Remark1.7.1.
In Section 1.5 and Section 1.6 we learned how to solve quadratic and rational equations. In this section, we use these skills to solve quadratic and rational inequalities.
Definition1.7.2.
A quadratic inequality is an inequality that can be written in one of the following forms:
\begin{equation*}
ax^2+bx+c \gt 0
\end{equation*}
\begin{equation*}
ax^2+bx+c \lt 0
\end{equation*}
\begin{equation*}
ax^2+bx+c \ge 0
\end{equation*}
\begin{equation*}
ax^2+bx+c \le 0
\end{equation*}
where \(a\text{,}\)\(b\text{,}\) and \(c\) are real numbers and \(a \neq 0\text{.}\)
Activity1.7.3.
Consider the graph shown below.
Figure1.7.4.
(a)
What is the value of \(f(0)\text{?}\)
\(\displaystyle 1\)
\(\displaystyle -4\)
\(\displaystyle 5\)
\(\displaystyle 0\)
Answer.
C
(b)
What is the value of \(f(3)\text{?}\)
\(\displaystyle 1\)
\(\displaystyle -4\)
\(\displaystyle 5\)
\(\displaystyle 0\)
Answer.
B
(c)
What is the value of \(f(5)\text{?}\)
\(\displaystyle 1\)
\(\displaystyle -4\)
\(\displaystyle 5\)
\(\displaystyle 0\)
Answer.
D
(d)
What is the value of \(f(6)\text{?}\)
\(\displaystyle 1\)
\(\displaystyle -4\)
\(\displaystyle 5\)
\(\displaystyle 0\)
Answer.
C
(e)
Notice from parts (a) - (d), that \(f(x)\) can either be positive, negative, or zero depending on the value of \(x\text{.}\) What are the \(x\)-intercept(s) of \(f(x)\text{?}\)
\(\displaystyle 1\)
\(\displaystyle -4\)
\(\displaystyle 5\)
\(\displaystyle 0\)
Answer.
A and C
(f)
Based on what you see on the graph (and your solutions to parts (a) - (d)), for what values of \(x\) would \(f(x)\) be positive?
\(\displaystyle x \lt 1\)
\(\displaystyle x \gt 1\)
\(\displaystyle x \gt 5\)
\(\displaystyle x \lt 5\)
\(\displaystyle 1 \lt x \lt 5\)
Answer.
A and C
(g)
Now use interval notation to express where \(x^2-6x+5 \gt 0\text{.}\)
\(\displaystyle (\infty, 1] \cup [5, \infty)\)
\(\displaystyle [1,5]\)
\(\displaystyle (\infty, 1) \cup (5, \infty)\)
\(\displaystyle (1,5)\)
Answer.
C
Remark1.7.5.
From Activity 1.7.3, we saw that a function could have \(y\)-values that are positive, negative, or zero, which can then help us find values of \(x\) to solve inequalities. Let’s now look at how we can solve inequalities using algebra.
Activity1.7.6.
Use Definition 1.7.2 and your knowledge of quadratic equations (see Section 1.5) to help answer the following questions.
(a)
Which of the following inequalities are quadratic inequalities?
\(\displaystyle (x-1)(x+3) \lt 7\)
\(\displaystyle -4x+3 \ge 10\)
\(\displaystyle 2x^2-7x+3 \gt 0\)
\(\displaystyle 5x-1 \le 4x\)
Answer.
A and C
(b)
Given the quadratic equation, \(x^2-x-6= 0\text{,}\) determine whether \(-1\) is a solution.
Answer.
By plugging in \(-1\) into the quadratic equation, \(x^2-x-6= 0\text{,}\) students should see that it is NOT a solution.
\((-1)^2-(-1)-6=-4\)
(c)
Given the quadratic inequality, \(x^2-x-6 \le 0\text{,}\) determine whether \(-1\) is a solution.
Answer.
By plugging in \(-1\) into the quadratic inequality, \(x^2-x-6 \le 0\text{,}\) students should see that it IS a solution as it produces a true statement.
\((-1)^2-(-1)-6=-4\) and \(-4 \le 0\text{.}\)
(d)
Are there any other values of \(x\) that would satisfy the inequality, \(x^2-x-6 \le 0\text{?}\)
Answer.
Any value of \(x\) from \(-2\) to \(3\) (including both \(-2\) and \(3\)) will satisfy the inequality. In interval notation, the solution is \([-2,3]\) (although students may not know this at this point in the section).
Remark1.7.7.
Notice from Activity 1.7.6, a solution to a quadratic inequality is a real number that will produce a true statement when substituted for the variable. Quadratic inequalities will often have an infinite number of solutions, which we will express in interval notation. However, it is also possible for the inequality to have no solution.
Activity1.7.8.
Let’s look at how we can algebraically determine the solutions to a quadratic inequality using a number line. Consider the quadratic inequality
What is the factored form of the quadratic (left-hand side of the inequality)?
\(\displaystyle (x-16)(x+2)\)
\(\displaystyle (x-4)(x+8)\)
\(\displaystyle (x+4)(x-8)\)
\(\displaystyle (x-4)(x-8)\)
Answer.
C
(b)
Rewrite the quadratic inequality with the answer you got from part (a). What values of \(x\) would give you \(0\text{?}\)
\(\displaystyle x = 16, -2\)
\(\displaystyle x = 4, -8\)
\(\displaystyle x = 4, 8\)
\(\displaystyle x = -4, 8\)
Hint.
Refer back to Section 1.5 and think about this as a quadratic equation (equal to \(0\)).
Answer.
D
(c)
These solutions (that you got in part (b)) correspond to all of the \(x\)-intercepts of the graph, and are the only spots where the \(y\)-values on either side might change from positive to negative or negative to positive. So, with our two \(x\)-intercepts, we have divided our graph into three intervals. What are these three intervals?
Answer.
\((-\infty, -4)\text{,}\)\((-4,8)\text{,}\) and \((8, \infty)\)
(d)
Notice that the \(x\)-intercepts are solutions to the quadratic equation \(x^2-4x-32=0\text{.}\) How should we mark these values of \(x\) on a number line?
Answer.
From Section 1.1, students should recall that because it is a greater than or equal to statement, that the \(x\)-intercepts should be included on the number (with a bracket or solid dot).
(e)
Choose a value of \(x\) within each interval and substitute that value into the equation \(x^2-4x-32\) (or the factored form you found in part (a)). If you get a positive value, place an "+" sign above that region on the number line. Similarly, if you get a negative value, place a "-" sign above that region on the number line.
Answer.
(f)
Using the number line and what you determined in part (e), shade in areas on the number line that satisfies the quadratic inequality \(x^2-4x-32 \ge 0\text{?}\)
Answer.
(g)
Using your graph, express the solution of the inequality \(x^2-4x-32 \ge 0\) in interval notation.
\(\displaystyle (-4,8)\)
\(\displaystyle [ -4,8]\)
\(\displaystyle (-\infty,-4)\cup(8,\infty)\)
\(\displaystyle (-\infty,-4]\cup[8,\infty)\)
Answer.
D
Remark1.7.9.
Notice in Activity 1.7.8, we had to begin with the solutions to the quadratic equation to determine regions of the number line to then test values of \(x\) that satisfy the inequality that was given. Creating a visual can help in determining the solutions to inequalities, especially when there are many solutions.
Definition1.7.10.
A sign chart is a number line representing the \(x\)-axis that shows where a function is positive or negative by using a ’+’ or a ’-’ sign to indicate which regions are positive or negative. For example, a sign chart for \(f(x)=x^2-4x-32\) (also seen in Activity 1.7.8) is shown below.
Figure1.7.11.A sign chart for the function \(f(x)=x^2-4x-32\text{.}\)
Definition1.7.12.
In Activity 1.7.8, we saw that when we place the solutions to the quadratic equation, \(x^2-4x-32=0\text{,}\) on the number line, that it divided the number line into three regions.
The values in the domain of a function that separate regions that produce positive or negative results are called critical points or boundary points. These values bound the regions where the function is positive or negative.
Activity1.7.13.
For this activity, consider the quadratic inequality
Factor the quadratic equation \(2x^2-28=10x\) using methods discussed in Section 1.5.
\(\displaystyle (x-2)(x+7)\)
\(\displaystyle (2x+4)(x-7)\)
\(\displaystyle (x+2)(x-7)\)
\(\displaystyle 2(x^2-14)\)
Answer.
B and C (although B is not completely factored)
(b)
Use the factors you found in part (a) to find the critical points.
\(\displaystyle x = -2, 7\)
\(\displaystyle x = \sqrt{14}\)
\(\displaystyle x = -7, 2\)
\(\displaystyle x = 2, -12\)
Answer.
A
(c)
Plot the critical points you found in part (b) on a number line. Determine whether those should be included in your solution based on the inequality that was given.
Answer.
Because the inequality is "less than", the critical points (or boundary points) should not be included in the solution.
(d)
Test values of \(x\) in each region that was created by the critical points and create a sign chart to show which regions are positive and negative.
Answer.
Figure1.7.14.
(e)
Using the number line and what you determined in part (d), shade in areas on the number line that satisfies the quadratic inequality \(2x^2-28 \lt 10x\text{?}\)
When solving quadratic inequalities, be sure to get all your terms to one side of the inequality first! Then, apply the methods we learned in Section 1.5 to determine the critical points (boundary points). From there, you can then create your sign chart to help determine the solution to the inequality.
Activity1.7.16.
For each of the following, determine the critical points and use a number line (and sign chart) to then find the solutions. Write your answers in interval notation.
\begin{equation*}
\frac{2x-3}{x-6}\le x
\end{equation*}
Activity1.7.19.
Use Definition 1.7.18 and your knowledge of quadratic equations (see Section 1.6) to help answer the following questions.
(a)
Which of the following inequalities are rational inequalities?
\(\displaystyle \frac{4x+7}{5} \lt -1\)
\(\displaystyle \frac{4x-1}{3x+2} \ge 0\)
\(\displaystyle 2x^2-7x \gt -3\)
\(\displaystyle 5x-1 \le 4x\)
Answer.
B
(b)
Given the rational equation, \(\frac{x-1}{x+3}= 0\text{,}\) determine whether \(1\) is a solution.
Answer.
By plugging in \(1\) into the rational equation, \(\frac{x-1}{x+3}= 0\text{,}\) students should see that it IS a solution.
\(\frac{1-1}{1+3}=\frac{0}{4} = 0\)
(c)
Given the rational equation, \(\frac{x-1}{x+3}= 0\text{,}\) determine whether \(-4\) is a solution.
Answer.
By plugging in \(-4\) into the rational equation, \(\frac{x-1}{x+3}= 0\text{,}\) students should see that it is NOT a solution.
\(\frac{-4-1}{-4+3}=\frac{-5}{-1}=5\text{,}\) which is NOT equal to \(0\text{.}\)
(d)
Given the rational inequality, \(\frac{x-1}{x+3} \ge 0\text{,}\) determine whether \(-4\) is a solution.
Answer.
By plugging in \(-4\) into the rational inequality, \(\frac{x-1}{x+3} \ge 0\text{,}\) students should see that it IS a solution.
\(\frac{-4-1}{-4+3}=\frac{-5}{-1}=5\text{,}\) which IS a solution to the inequality because it is greater than \(0\text{.}\)
(e)
Given the rational equation, \(\frac{x-1}{x+3}= 0\text{,}\) determine whether \(-3\) is a solution.
Answer.
By plugging in \(-3\) into the rational inequality, \(\frac{x-1}{x+3} \ge 0\text{,}\) students should see that it is NOT a solution.
\(\frac{-3-1}{-3+3}=\frac{-4}{0}\text{,}\) which is undefined.
(f)
Are there any other values of \(x\) that would satisfy the inequality, \(\frac{x-1}{x+3} \ge 0\text{?}\)
Answer.
Any value of \(x\) less than \(-3\) or greater than or equal to \(1\) will satisfy the inequality. In interval notation, the solution is \((-\infty, -3) \cup [1, \infty)\) (although students may not know this at this point in the section).
Remark1.7.20.
In Activity 1.7.19, we saw that there can be solutions to the rational inequality \(\frac{x-1}{x+3} \ge 0\) that does NOT satisfy the rational equation \(\frac{x-1}{x+3}= 0\text{.}\) Just like with linear and quadratic inequalities, we can have many solutions that can satisfy rational inequalities. We do, however, need to be careful about our critical points, as we will see in the next activity.
Focus on the rational equation \(\frac{x-1}{x+3}\) (the left-hand side of the inequality). What value(s) should be excluded as possible solutions to that rational expression?
\(\displaystyle 1\)
\(\displaystyle -1\)
\(\displaystyle -3\)
\(\displaystyle 3\)
Answer.
C
(b)
The value you got in part (a) is a critical value for this rational expression because it is a value of \(x\) where the rational expression is undefined. Critical points for rational expressions also include values that make the rational expression equal to \(0\text{.}\) What value of \(x\) would make this rational expression equal to \(0\text{?}\)
\(\displaystyle 1\)
\(\displaystyle -1\)
\(\displaystyle -3\)
\(\displaystyle 3\)
Answer.
A
(c)
Use the critical points you found in parts (a) and (b) and plot them on a number line. Determine whether those should be included in your solution based on the inequality that was given.
Answer.
The value \(1\) IS a solution to the inequality because although it makes the rational expression equal to \(0\text{,}\) the inequality states "greater than or equal to \(0\text{.}\)" The other critical value, \(-3\text{,}\) however should be in the solution as it makes the rational expression undefined. So, on a number line, \(-3\) should have a parentheses and \(1\) should have a bracket.
(d)
Notice that your number line has now been divided into three regions. Test values of \(x\) in each region that was created by the critical points and create a sign chart to show which regions are positive and negative.
Hint.
Make sure you substitute values into BOTH \(x\)s in the rational expression.
Answer.
Figure1.7.22.
(e)
Using the number line and what you determined in part (d), shade in areas on the number line that satisfies the quadratic inequality \(\frac{x-1}{x+3} \ge 0\text{?}\)
When looking for critical points of rational inequalities, remember to look for points where the rational expression will be zero or undefined.
Activity1.7.24.
For each of the following, determine the critical points and use a number line (and sign chart) to then find the solutions. Write your answers in interval notation.
(a)
\(\frac{x+5}{x-4} \le 0\)
Answer.
\([-5,4)\)
(b)
\(\frac{(x+3)(x+5)}{x+2} \gt 0\)
Hint.
Notice for this rational inequality there are \(3\) critical points! Be sure to test all four regions on your number line.
Answer.
\((-5,-3)\)
Activity1.7.25.
When solving an inequality, the goal is to first get \(x\) (or whatever the variable is) on its own on one side of the inequality sign and \(0\) on the other side. To do this, we have to be careful of the actions we take as some actions can change the direction of the inequality. Let’s revisit some of the actions we have taken previously and see how we can apply these same actions to solve the rational inequality
When solving rational equations in Section 1.6, we often started by "clearing the fractions" by multiplying the denominator to both sides of the equation. Suppose we have the equation \(\frac{3x-10}{x-4} = 2\text{.}\) What can we multiply each term by that will clear the fraction?
\(\displaystyle x+4\)
\(\displaystyle 3x-10\)
\(\displaystyle x-4\)
\(\displaystyle x-2\)
Answer.
C
(b)
Multiply each term by the expression you chose and simplify. Which of the following linear equations does the rational equation simplify to?
\(\displaystyle (x+4)(3x-10)=2(x+4)\)
\(\displaystyle (3x-10)=2(x-4)\)
\(\displaystyle 2(3x-10)=(x-4)\)
\(\displaystyle (3x-10)=2(x+4)\)
Answer.
B
(c)
What values of \(x\) would make the denominator of the rational expression positive?
Answer.
Any value of \(x\) greater than or equal to \(4\text{.}\)
(d)
What values of \(x\) would make the denominator of the rational expression negative?
Answer.
Any value of \(x\) less than \(4\text{.}\)
(e)
Now suppose we applied the same actions as we did in parts (a) and (b) to the rational inequality \(\frac{3x-10}{x-4} \gt 2\text{.}\) If the denominator of the rational expression \(\frac{3x-10}{x-4}\) was positive, how would you write the inequality as a linear inequality?
\(\displaystyle (x+4)(3x-10)\gt 2(x+4)\)
\(\displaystyle (3x-10) \gt 2(x-4)\)
\(\displaystyle 2(3x-10) \gt(x-4)\)
\(\displaystyle (3x-10) \gt (x+4)\)
Answer.
B
(f)
If the denominator of the rational expression \(\frac{3x-10}{x-4}\) was negative, how would you write the inequality as a linear inequality?
\(\displaystyle (x+4)(3x-10)\lt 2(x+4)\)
\(\displaystyle (3x-10) \gt 2(x-4)\)
\(\displaystyle 2(3x-10) \lt(x-4)\)
\(\displaystyle (3x-10) \lt 2(x-4)\)
Hint.
What happens to the inequality symbol when multiplying or dividing by a negative number?
Answer.
D
(g)
Look at your answers in parts (e) and (f). Which inequality would help us solve \(\frac{3x-10}{x-4} \gt 2\text{.}\)
Answer.
Students should see that it is not clear which linear inequality to use because there are values of \(x\) that can make the denominator of the rational expression positive or negative.
Remark1.7.26.
In Activity 1.7.25, we saw that it is not clear whether the denominator of a rational expression would yield a positive or negative value. Because we do not actually know whether the denominator is positive or negative, we cannot multiply the denominator to "clear the fractions" as we did before when solving rational equations. In the next activity, we will look at how we can solve rational inequalities.
From Activity 1.7.25, we saw that we cannot multiply by the denominator to "clear the fraction." Our goal, however, is still to get all the \(x\)s to one side and \(0\) on the other side. What action can we take to get \(0\) on one side?
Add \(2\) to each side
Subtract \(2\) from each side
Multiply by \(2\) on each side
Multiply by \(4\) on each side
Answer.
B
(b)
What rational inequality do you now have?
\(\displaystyle \frac{3x-10}{x-4}+2 \lt 0\)
\(\displaystyle \frac{3x-10}{x-4}-2 \lt 0\)
\(\displaystyle \frac{3x-10}{x-4}+2 \gt 0\)
\(\displaystyle \frac{3x-10}{x-4}-2 \gt 0\)
Answer.
D
(c)
What is the common denominator of \(\frac{3x-10}{x-4}\) and \(2\text{?}\)
\(\displaystyle x+4\)
\(\displaystyle x-4\)
\(\displaystyle 3x-10\)
\(\displaystyle 3x+10\)
Answer.
B
(d)
Multiply \(2\) by the common denominator and simplify the rational expression. What rational inequality do you now have?
\(\displaystyle \frac{x-18}{x-4} \gt 0\)
\(\displaystyle \frac{x-2}{x-4} \gt 0\)
\(\displaystyle \frac{x-2}{x-4} \gt 0\)
\(\displaystyle \frac{2x-6}{x-4} \gt 0\)
Answer.
C
(e)
Use the inequality you got in part (d) to determine the critical points.
Hint.
Remember that for a rational inequality, the critical points are the values of \(x\) that make the rational expression equal to \(0\) or undefined.
Answer.
\(2\) and \(4\)
(f)
Use the critical points you found in part (e) and plot them on a number line. Determine whether those should be included in your solution based on the inequality that was given.
Answer.
Both values of \(x\) should be excluded from the solution. On the number line, they would have parentheses (or open circles).
(g)
Use the critical points to create regions on the number line for you to test values of \(x\) and create a sign chart to show which regions are positive and negative.
Answer.
Figure1.7.28.
(h)
Using the number line and what you determined in part (d), shade in areas on the number line that satisfies the rational inequality \(\frac{3x-10}{x-4} \gt 2\text{?}\)
Subtract \(x\) on both sides of the inequality to get \(0\) on one side. Simplify \(\dfrac{4x+3}{x+2} - x\) into a single rational expression.
\(\displaystyle \dfrac{4x+3}{x+2}\)
\(\displaystyle \dfrac{3x+3}{x+2}\)
\(\displaystyle \dfrac{x^2+6x+3}{x+2}\)
\(\displaystyle \dfrac{-x^2+2x+3}{x+2}\)
Answer.
D
(b)
What are the critical points of this rational inequality?
Hint.
Factor the numerator.
Answer.
\(x = -2, -1, 3\)
(c)
Use the critical points you found in part (b) and plot them on a number line. Determine whether those should be included in your solution based on the inequality that was given.
Answer.
All critical points are not included in the solution.
(d)
Use the critical points to create regions on the number line for you to test values of \(x\) and create a sign chart to show which regions are positive and negative.
Answer.
Figure1.7.30.
(e)
Using the number line and what you determined in part (d), shade in areas on the number line that satisfies the rational inequality \(\dfrac{4x+3}{x+2} \gt x\text{?}\)
Answer.
(f)
How can we express the answers to part (e) for the rational inequality using interval notation?
For each of the following, solve the rational inequality by bringing all \(x\) terms to one side (and \(0\) on the other) and simplify the rational expression. Then, use critical points and the sign chart/number line to determine the solution.
