In today’s blog I will show 3 different ways for solving a quadratic inequality, including one way that relies on the conceptual understanding of parabolas.

Suppose you want to solve the inequality $x^2+x-20\le 0$.

Critical Values

The first step is to find the critical values associated with this inequality. These are the values of x for which $x^2+x-20$ is equal to 0. In this case the equation can be solved by factoring, but keep in mind that some equations will require the use of the quadratic formula or another method.

$x^2+x-20=0\\(x+5)(x-4)=0\\ \begin{array}{rcl} x+5=0 & or & x-4=0 \end{array}\\ \begin{array}{rcl} x=-5 & or & x=4 \end{array}$

So, the critical values are $\begin{array}{rcl} x=-5 & and & x=4 \end{array}$. Place those two values on a number line, which breaks the number line into 3 intervals as shown.

Test Points

Now we select one test point from each interval to determine which intervals contain solutions. In this example we could select -6 from interval I, 0 from interval II, and 5 from interval III. (Those selections are arbitrary, we can select any value from each interval.) There are three options for testing the intervals.

Option 1 – Substitute the values

We can substitute each value into the original inequality and determine whether it is a solution or not.

$\begin{array}{ccc} TP & x^2+x-20 & \le 0? \\ -6 & (-6)^2+(-6)-20=10 & False \\ 0 & (0)^2+(0)-20=-20 & True \\ 5 & (5)^2+(5)-20=10 & False \end{array}$

The only test point that was a solution is $x=0$, so interval II contains the solutions to this inequality.

The disadvantage to this method is that students are prone to making simple arithmetic mistakes.

Option 2 – Create a sign chart

Instead of substituting into the original inequality, another option is to create a sign chart. This technique can only be used for inequalities comparing an expression to 0 when the quadratic expression can be factored.
The sign chart focuses on the sign of each factor to determine whether the entire expression is positive $(>0)$ or negative $(<0)$.
Here’s the sign chart for this example – we are looking for the expression $x^2+x-20$ to be negative because the original inequality was $x^2+x-20 \le 0$.

$\begin{array}{cccc} TP & x+5 & x-4 & x^2+x-20 \\ -6 & - & - & + \\ 0 & + & - & - \\ 5 & + & + & +\end{array}$

For $x=-6$, $x+5$ is equal to -1, so we write a negative sign under $x+5$.
The second factor, $x-4$ is equal to -10 for $x=-6$, so we write a negative sign under that factor as well.
The product of two negative numbers is positive, so we write a positive sign under $x^2+x-20$.

Since the test point $x=0$ is the only test point that produced a negative result, interval II contains the solutions.

Option 3 – Think about the graph of the associated parabola

If we were to graph the equation $y=x^2+x-20$, we would have a parabola opening upward whose x-intercepts are located at the critical values of x. The vertex would be located below the x-axis between those two x-intercepts.

The intervals for which the graph is above the x-axis correspond to intervals for which $x^2+x-20>0$, because the y-coordinates are positive.
The intervals for which the graph is below the x-axis correspond to intervals for which $x^2+x-20<0$. because the y-coordinates are negative.

In this example, the graph is below the x-axis between -5 and 4, so the solutions can be found in interval II.

You do not need to graph the parabola to be able to use this approach. If you can picture the graph of the parabola in your head (does it open up or down), then you can also visualize the intervals that the graph is below the x-axis and the intervals that the graph is above the x-axis.

Solution

I recommend that students put their solution on a number line, and then write the interval notation associated with the solutions on the number line. Here is the number line showing the solutions.

In interval notation, this can be expressed as $[-5,4]$.

(By the way, I graphed the parabola and the number line at webgraphing.com. They have a fantastic free graphing site there! I’ll share more in a future blog.)

Summary

I hope that you find this helpful. If you have any questions about solving quadratic inequalities, or if you have any other topics you are interested in, you can reach me through the contact page at my web site – georgewoodbury.com.

-George

I am a math instructor at College of the Sequoias in Visalia, CA. Each Wednesday I post an article related to General Teaching on my blog. If there’s a particular topic you’d like me to address, or if you have a question or a comment, please let me know. You can reach me through the contact page on my website – http://georgewoodbury.com

• 1. Dave  |  October 27, 2010 at 4:19 pm

A single test point should suffice when solving a rational inequality. Once we know the sign in one interval, we can quickly deduce the signs in the other intervals, because the sign changes at x=c when (x-c) occurs to an odd power in the numerator or denominator, and it remains the same when (x-c) occurs to an even power.

I have tried to teach students this method. Sometimes they catch on, but they seem to find it confusing, or they resist it because it’s unfamiliar. They seem to prefer choosing one test point in each interval, even though it seems to be a lot of extra work.

• 2. georgewoodbury  |  November 3, 2010 at 5:46 am

Dave – Good point. When we teach this again in Precalculus and beyond, we definitely make use of this. Actually, I start with end behavior to determine the sign of the expression in the last interval and work backwards from there.

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• 4. t42ng  |  November 29, 2011 at 10:55 pm

Using 3 test points for all areas on the number line is a good lesson to students who learn solving quadratic inequalities for the first time. After that, applying notion of symmetry, taking the origin (x = 0) as unique test point is enough. During tests/exams time is very limited, there is not enough time for taking 3 tests points.
There is an European algebraic method that is based on a theorem:”Between the 2 real roots, the trinomial f(x) has the opposite sign of the coefficient a”. This means f(x) > 0 when a is negative, and vice versa. Students learn the theorem’s development once then apply it for solving quadratic inequalities. For multiple choice answers in test/exam, by this algebraic method, students can immediately give the answer without having to draw the number line each time.
Example. Solve f(x) = x^2 – 6x – 7 < 0. The 2 real roots are -1 and 7. Since a is positive, f(x) < 0 between -1 and 7. The answer is the open interval (-1, 7).

• 5. t42ng  |  November 29, 2011 at 11:13 pm

Sorry for the mistake about the above example. Please correct the equation: f(x) = x^2 – 6x – 7 < 0
Example 2. Solve : x^2 – 5x + 6 < 0. The 2 real roots are -1 and 6. Since a is positive, f(x) 0. The 2 real roots are 1 and 7. Since a is negative, f(x) > 0 between 1 and 7. The solution set is the open interval (1, 7).
There is no need for drawing the number line !

• 6. t42ng  |  November 29, 2011 at 11:26 pm

Example 2. Solve: f(x) = x^2 – 5x – 6 < 0. The 2 real roots are -1 and -6. Since a is positive, f(x) 0. The 2 real roots are 1 and -12.. Since a is negative, f(x) > 0 between -12 and 1. The solution set is the open interval (-12, 1).
There is no need for drawing a number line!