2.5: Polynomial and Rational Inequalities
Learning Objectives
 Understand the properties of quadratic inequalities.
 Understand the properties of polynomial inequalities.
 Understand the properties rational inequalities.
 Apply understanding of quadratic, polynomial, and rational inequalities to draw graphs.
Quadratic Inequalities
Quadratic inequalities are inequalities that have one of the following forms
\begin{align*}ax^{2}+bx+c>0\end{align*}
and
\begin{align*}ax^{2}+bx+c<0\end{align*}
We can solve these inequalities by using the techniques that we have learned about solving quadratic equations. For example, consider the graph of the equation:
\begin{align*}y=f(x)=x^{2}+x6\end{align*}
Notice that the curve intersects the \begin{align*}x\end{align*}axis at 3 and 2. From graph, we notice the followings
 If \begin{align*}x< 3\end{align*} then \begin{align*}f(x)>0\end{align*}
 If \begin{align*}3<x<2\end{align*}, then \begin{align*}f(x)<0\end{align*}
 If \begin{align*}x>2\end{align*}, then \begin{align*}f(x)>0\end{align*}
Therefore, \begin{align*}x^{2}+x6>0\end{align*} whenever \begin{align*}x<3\end{align*} or \begin{align*}x>2\end{align*}, and \begin{align*}x^{2}+x6<0\end{align*} when \begin{align*}3<x<2\end{align*}.
Example 1
What is the solution set of the inequality \begin{align*}2x^{2}+7x4<0\end{align*}?
Solution
It is best to graph the function \begin{align*}f(x)=2x^{2}+7x4\end{align*} and look for the the values of \begin{align*}x\end{align*} such that the inequality \begin{align*}f(x)<0\end{align*} is true.
Thus from graph, \begin{align*}2x^{2}+7x4<0\end{align*} only if
\begin{align*}4<x<\frac{1}{2}\end{align*}
So the solution set is \begin{align*}x\in\left ( 4,\frac{1}{2} \right )\end{align*} or in set builder notation, \begin{align*}\left \{x4<x<\frac{1}{2} \right \}\end{align*}.
Although the method of graphing to find the solution set of an inequality is easy to follow, another algebraic method can be used. The algebraic method involves finding the \begin{align*}x\end{align*}intercepts of the graph and then dividing the \begin{align*}x\end{align*}axis into intervals separated by the \begin{align*}x\end{align*}intercepts. The examples below illustrate the method.
Example 2
Find the solution set of the quadratic inequality \begin{align*}x^{2}+2x8>0\end{align*}.
Solution
As you know the graph of the function will immediately tell us at which intervals the inequality is positive. However, we will not draw a graph and find the solution without its aid. Rather, we solve
\begin{align*}x^{2}+2x8 & = 0\\ (x+4)(x2) & = 0\end{align*}
The two solutions to this quadratic equation are \begin{align*}x=4\end{align*} and \begin{align*}x=2\end{align*}, thus, the \begin{align*}x\end{align*}intercepts of the function \begin{align*}f(x)=x^{2}+2x8\end{align*} are 4 and 2. These points divide the \begin{align*}x\end{align*}axis into three intervals: \begin{align*}(\infty,4), (4,2)\end{align*} and \begin{align*}(2,\infty)\end{align*}. We can choose a test point from each interval and substitute it into \begin{align*}f(x)\end{align*} and see if it is negative or positive. This procedure can be simplified by making the table shown below.
Interval  Test Point  \begin{align*}x^{2}+2x8\end{align*} Positive/Negative?  Part of Solution set? 

\begin{align*}(\infty,4)\end{align*}  5  +  yes 
(4, 2)  1    no 
\begin{align*}(2,+\infty)\end{align*}  3  +  yes 
From the table, we conclude that since \begin{align*}x^{2}+2x8>0\end{align*} if and only if \begin{align*}x<4\end{align*} and \begin{align*}x>2\end{align*}. The solution set can also be written as
\begin{align*}x\in(\infty,4)\cup(2,+\infty)\end{align*}
Some problems in science involve quadratic inequalities. The example below illustrates one such application.
Example 3
A rectangle has a length 10 meters more than twice the width. Find the possible widths such that the area of the rectangle can not exceed 100 squared meters.
Solution
Let \begin{align*}W\end{align*} be the width of the rectangle and \begin{align*}L\end{align*} its length. Thus
\begin{align*}L & = 10+2W\\ W&=W\end{align*}
The area of a rectangle is
\begin{align*}A & = LW\\ & = (10+2W)(W)\\ & = 10W +2W^2\end{align*}
We are requiring that the area can not exceed \begin{align*}100 \ m^{2}\end{align*}. So
\begin{align*}10W+2W^{2}<100\end{align*}
or
\begin{align*}2W^{2}+10W100<0\end{align*}
Dividing both sides by 2,
\begin{align*}W^{2}+5W50<0\end{align*}
Finding the roots for \begin{align*}W^{2}+5W50=0\end{align*},
\begin{align*}W^{2}+5W50=(W+10)(W5)\end{align*}
So the partition points are 5 and 10 and thus we have three intervals. Since width can not be negative, we can safely ignore 10. The maximum area is \begin{align*}100 \ m^{2}\end{align*} and thus the width must be \begin{align*}W<5\end{align*}.
Solve Polynomial Inequalities
Solving polynomial inequalities is very similar to solving quadratic inequalities. The basic steps are the same:
 Set up the inequality in the form \begin{align*}p(x)>0\end{align*} (or \begin{align*}p(x)<0, p(x)\le0,p(x)\ge0\end{align*})
 Find the solutions to the equation \begin{align*}p(x)=0\end{align*}.
 Divide the number line into intervals based on the solutions to \begin{align*}p(x)=0\end{align*}.
 Use test points to find solution sets to the equation.
Example 4
Solve \begin{align*}x^{3}3x^{2}+2x\ge0\end{align*}
Solution
The polynomial is already in the correct form \begin{align*}p(x)\ge0\end{align*} so we solve the equation
\begin{align*}x^{3}3x^{2}+2x & = 0\\ x(x^23x+2) & = 0\\ x(x2)(x1) & = 0\end{align*}
The zeros are at \begin{align*}x=0, x=1,\end{align*} and \begin{align*}x=2\end{align*}.
Interval  Test Point  Positive/Negative?  Part of Solution set? 

\begin{align*}(\infty, 0)\end{align*}  5    no 
(0, 1)  \begin{align*}\frac{1}{2}\end{align*}  +  yes 
(1, 2)  \begin{align*}\frac{3}{2}\end{align*}    no 
\begin{align*}(2,+\infty)\end{align*}  3  +  yes 
Notice that this inequality is greater than or equal to zero, so we include the zeros in the solution set. Therefore the solutions are \begin{align*}x\in[0,1]\cup[2,+\infty]\end{align*}.
Example 5
Solve \begin{align*}6x^{4}+20x^{2}<25\end{align*}.
Solution
First we will change the inequality to \begin{align*}6x^{4}+20x^{2}25<0\end{align*}. Now, solve the equation \begin{align*}6x^{4}+20x^{2}25=0\end{align*}.
\begin{align*}6x^{4}+20x^{2}25 & = 0\\ (3x^25)(2x^2+5) & = 0\end{align*}
The first term yields the solutions \begin{align*}x=\pm\sqrt{\frac{5}{3}}\end{align*} and there are no real solutions for the second term.
Interval  Test Point  Positive/Negative?  Part of Solution set? 

\begin{align*}\left ( \infty,\sqrt{\frac{5}{3}} \right )\end{align*}  3  +  no 
\begin{align*}\left ( \sqrt{\frac{5}{3}},\sqrt{\frac{5}{3}} \right )\end{align*}  0    yes 
\begin{align*}\left ( \sqrt{\frac{5}{3}},+\infty \right )\end{align*}  3  +  no 
Finally, the solution set is \begin{align*}x\in\left ( \sqrt{\frac{5}{3}},\sqrt{\frac{5}{3}} \right )\end{align*}.
Notice that all of the polynomials we have picked so far are easily factorable. At this point the set of polynomial inequalities you can solve using algebraic methods is fairly limited. Later in this section we look at how a graphing calculator can help you solve polynomial inequalities. Also, once you learn more advanced methods for solving polynomial equations and finding rational zeros, you can apply those methods to solving different types of polynomial inequalities.
Solve Rational Inequalities
The method outlined above also works for solving inequalities involving rational functions. Recall that for rational functions you can find the roots (or zeros) by setting the numerator equal to zero.
However there is one step added to the process of solving rational inequalities because a rational function can also change signs at its vertical asymptotes or at a break in the graph. For instance, look at the graph of the function \begin{align*}r(x)=\frac{x}{x^{2}9}\end{align*} below.
If we want to solve the inequality \begin{align*}\frac{x}{x^{2}9}>0\end{align*}, then we need to use the following critical points: \begin{align*}x=0, x=3,\end{align*} and \begin{align*}x=3\end{align*}. \begin{align*}x=0\end{align*} is the solution of setting the numerator equal to 0, and this gives us the only root of the function. \begin{align*}x=\pm3\end{align*} are the vertical asymptotes, the \begin{align*}x\end{align*}coordinates that make the function undefined because putting in 3 for \begin{align*}x\end{align*} will cause a division by zero.
Using the graph or test points, we can build the table,
Interval  Test Point  Positive/Negative?  Part of Solution set? 

\begin{align*}(\infty,3)\end{align*}  4    no 
(3, 0)  2  +  yes 
(0, 3)  2    no 
\begin{align*}(3,+\infty)\end{align*}  4  +  yes 
Thus, the solutions to \begin{align*}\frac{x}{x^{2}9}>0\end{align*} are \begin{align*}x\in(3,0)\cup(3,+\infty)\end{align*}.
Example 6
Find the solution set of the inequality
\begin{align*}\frac{4x12}{3x2}<0\end{align*}
From the numerator we solve \begin{align*}4x12=0\end{align*} or \begin{align*}x=3\end{align*}. In the denominator, solve \begin{align*}3x2=0\end{align*} and we find the critical point \begin{align*}x=\frac{2}{3}\end{align*}.
Making the table
Interval  Test Point  Positive/Negative?  Part of Solution set? 

\begin{align*}\left ( \infty,\frac{2}{3} \right )\end{align*}  0  +  no 
\begin{align*}\left ( \frac{2}{3},3 \right )\end{align*}  1    yes 
\begin{align*}(3,+\infty)\end{align*}  5  +  no 
Therefore, the solution set includes the numbers in the interval \begin{align*}\left ( \frac{2}{3},3 \right )\end{align*}. Or in setbuilder notation, the solution is \begin{align*}\left \{ x\frac{2}{3}<x<3 \right \}\end{align*}.
Summarize Polynomial and Rational Inequalities
The basic method for solving polynomial and rational inequalities is the same:
 Rewrite the inequality so that it is in terms of zero.
 Find the zeros and critical points to divide the domain into intervals
 Use test points in each interval to see which intervals satisfy the inequality
 Build your solution set from the table of intervals and test points
Example 7
Solve the inequality \begin{align*}\frac{3x+2}{x^{2}}>3\end{align*}
Solution
First, we use algebra to rewrite the inequality to get zero on one side
\begin{align*}\frac{3x+2}{x^{2}} & > 3\\ \frac{3x+2}{x^{2}}  3 & > 0\\ \frac{3x^2+3x+2}{x^{2}} & > 0\\ \frac{3x^23x2}{x^{2}} & < 0\end{align*}
Notice in the last step we multiplied both sides by 1, so we changed the direction of the inequality. The numerator cannot be factored, so we use the quadratic formula to solve \begin{align*}3x^{2}3x2=0\end{align*}.
\begin{align*}x & = \frac{ 3 \pm \sqrt{94(3)(2)}}{6}\\ x & = \frac{3 \pm \sqrt{33}}{6}\end{align*}
So the two zeros of the rational function are \begin{align*}x=\frac{3+\sqrt{33}}{6}\approx 1.457\end{align*} and \begin{align*}x=\frac{3\sqrt{33}}{6}\approx0.457\end{align*}. The final critical point is \begin{align*}x=0\end{align*} because of the \begin{align*}x^{2}\end{align*} term in the denominator of the inequality. Using these values we construct the table:
Interval  Test Point  Positive/Negative?  Part of Solution set? 

\begin{align*}\left ( \infty,\frac{3\sqrt{33}}{6} \right )\end{align*}  1  +  no 
\begin{align*}\left ( \frac{3\sqrt{33}}{6},0 \right )\end{align*}  \begin{align*}\frac{1}{4}\end{align*}    yes 
\begin{align*}\left ( 0,\frac{3+\sqrt{33}}{6} \right )\end{align*}  1    yes 
\begin{align*}\left ( \frac{3\sqrt{33}}{6},+\infty \right )\end{align*}  2  +  no 
The final solution set is \begin{align*}x\in\left ( \frac{3\sqrt{33}}{6},0 \right )\cup\left ( 0,\frac{3+\sqrt{33}}{6} \right )\end{align*}
Notice in this example that the intervals of the solution are on both sides of \begin{align*}x=0\end{align*}. You may be tempted to include \begin{align*}x=0\end{align*} in the solution set, but that will not work. The reason is because at \begin{align*}x=0\end{align*} the rational function is undefined, so 0 cannot satisfy the inequality.
Applications, Technological Tools
Solving quadratic, polynomial, and rational inequalities is much easier with a calculator. Two specific functions that a TI83/84 calculator provides to help solve rational inequalities are:
 Using the calculator to graph a function and using the CALC menu to identify its roots.
 Using the table function to substitute test values into the function.
In a previous section we outlined stepbystep directions for doing both of these tasks. Here we will look at one example that brings these steps together to solve a polynomial inequality that could not be done using simple algebraic methods.
Example 8
The McNeil Surf Company makes wetsuits. For a given number of wetsuits \begin{align*}x\end{align*}, McNeil's profit, in dollars, is given by the function \begin{align*}P(x)=0.01x^{2}+25x3000\end{align*}.
a) If the manager of McNeil wants the profit to stay above $9,000, what is the minimum and maximum number of wetsuits they can manufacture to maintain that level of profit?
b) What is the maximum profit McNeil can make?
c) Can you explain why this shape might make sense for a profit function?
Solution
a) Set up the inequality
\begin{align*}0.01x^{2}+25x3000 & > 9000\\ 0.01x^2 + 25x 12000 & > 0\end{align*}
With a calculator you can graph the function \begin{align*}Y_{1}=0.01x^{2}+25x12000\end{align*}. Here is the graph using a window \begin{align*}[1000,4000] \times [5000,15000]\end{align*}.
((INSERT TI83 Graph with WINDOW, \begin{align*}Xmin = 1000, Xmax = 4000, Xscl = 500, Ymin=5000, Ymax=5000, Yscl=1000 \ xres=1\end{align*}))
((INSERT TI83/84 Graph of the function))
Using the CALC menu (2ND TRACE), and selecting the option ZEROS, we can see that the zeros of \begin{align*}Y_{1}=0.01x^{2}+25x12000\end{align*} are at \begin{align*}x=647.920\end{align*} and \begin{align*}x=1852.080\end{align*}. By inspecting the graph, we can see that the solution set to the inequality \begin{align*}0.01x^{2}+25x12000>0\end{align*} is \begin{align*}x\in(648,1852)\end{align*}.
b) Keeping the same graph open, use CALC MAXIMUM to solve for the maximum profit. The maximum is at (1250, 3625), indicating that the maximum profit is $3625 when 1250 wetsuits are produced.
c) One possible reason the profit function might take this shape is labor costs. If McNeil wants to make a very large number of wetsuits in a short period of time, then that may require paying overtime for workers, and this could reduce the profit margin.
Exercises
 Find the solution set of the following inequalities without using a calculator. Display the solution set on the number line.
 \begin{align*}x^{2}+2x3\le0\end{align*}
 \begin{align*}3x^{2}7x+2>0\end{align*}
 \begin{align*}6x^{2}13x+5\ge0\end{align*}
 \begin{align*}\frac{5x1}{x2}>0\end{align*}
 \begin{align*}\frac{1x}{x}<1\end{align*}
 Use a calculator to solve the following inequalities. Round your answer three places after the decimal.
 \begin{align*}9.8t^{2}+357.6t\ge0\end{align*}
 \begin{align*}x^{3}5x+7\le4x^{2}+18\end{align*}
 \begin{align*}\frac{x^{2}2x}{x5}>x^{2}25\end{align*}
 The total resistance of two electronics components wired in parallel is given by \begin{align*}\frac{R_{1}R_{2}}{R_{1}+R_{2}}\end{align*} where \begin{align*}R_{1}\end{align*} and \begin{align*}R_{2}\end{align*} are the individual resistances (in Ohms) of the two components.
 If the resistance of \begin{align*}R_{1}\end{align*} is 20 Ohms, what is the maximum resistance of \begin{align*}R_{2}\end{align*} if the total resistance must be less than 15 Ohms?
 What is the maximum theoretical resistance of this circuit? How do you know?
 A rectangular lot of land has a length that is 7 meters more than twice its width. If the area of the lot is greater than 60 square meters, what are the possible values of the widths of the lot?
Answers

 \begin{align*}x\in[3,1]\end{align*}
 \begin{align*}x\in\left ( \infty,\frac{1}{3} \right )\cup(2,+\infty)\end{align*}
 \begin{align*}x\in\left [ \frac{5}{2},\frac{1}{3} \right ]\end{align*}
 \begin{align*}x\in\left ( \infty,\frac{1}{5} \right )\cup(2,+\infty)\end{align*}
 \begin{align*}x\in(\infty,0)\cup\left ( \frac{1}{2},+\infty \right )\end{align*}
 \begin{align*}t\in[0,36.490]\end{align*}
 \begin{align*}x\in(\infty,4.567]\cup[1.294,1.861]\end{align*}
 \begin{align*}x\in(4.667,4.044)\cup(5,6.623)\end{align*}
 60 Ohms
 The total resistance will always be less than 20 because \begin{align*}y=20\end{align*} is the horizontal asymptote of \begin{align*}y=\frac{20R_{2}}{20+R_{2}}\end{align*}.
 Width must be greater than 4 meters, \begin{align*}w>4\end{align*}.
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