This section covers:
 Graphing Quadratic Inequality Functions
 Solving Quadratic Inequalities
 Solving Using Graphing
 Solving Algebraically, including Completing the Square
 Sign Chart (Sign Pattern) Method
 Real World Quadratic Inequality
 More Practice
Just like we solved and graphed Linear Inequalities, we can do the same with Quadratic Inequalities.
(See the Solving Inequalities Section for information on other inequalities.)
Graphing Quadratic Inequality Functions
We learned how to graph inequalities with two variables way back in the Coordinate System and Graphing Lines section. We can do the same with quadratics and the shading is pretty much the same: when we have “\(y<\) ”, we always shade in under the line that we draw, and when we have “\(y>\) ”, we always shade above the line that we draw. We can even do these on a graphing calculator!
Again, we can always plug in an ordered pair to see if it shows up in the shaded areas (which means it’s a solution), or the unshaded areas (which means it’s not a solution.) With “\(<\)” and “\(>\)” inequalities, we draw a dashed (or dotted) line to indicate that we’re not really including that line (but everything up to it), whereas with “\(\le \)” and “\(\ge \)”, we draw a regular line, to indicate that we are including it in the solution. To remember this, I think about the fact that “\(<\)” and “\(>\)” have less pencil marks than “\(\le \)” and “\(\ge \)”, so there is less pencil used when you draw the lines on the graph. You can also remember this by thinking the line under the “\(\le \)” and “\(\ge \)” means you draw a solid line on the graph.
Quadratic Inequality/Explanation  Graph 
\(y<5{{x}^{2}}3x+2\)
First, graph the parabola \(y=5{{x}^{2}}3x+2\). Since we have “\(y<\)” inequality, we shade under the graph, since it “rains down”. Note that we use a dashed line since we have a “\(<\)”.
If we had “\(>\)”, it would be shaded up (inside the parabola).
Let’s plug in \(\left( {0,0} \right)\) to see if it shows up as a solution. \(\left( {0,0} \right)\) is in the shaded area, so it satisfies the inequality: \(0<5{{\left( 0 \right)}^{2}}3\left( 0 \right)+2\). 

We can also graph this using a Graphing Calculator.
After typing in the equation in \({{Y}_{1}}=\), use the cursor to scroll over to before the \({{Y}_{1}}=\) and keep hitting ENTER until you see the. Then hit GRAPH.
(For the TI84C Calculator, you’ll have to use the curser to scroll down past the Color: setting to Y, and use the scroll right curser to get to “\(<\)”.) 

Solving Quadratic Inequalities
You will also have to know how to solve quadratic inequalities, which make things a little messy. Examples of Quadratic Inequalities can be as “simple” as \({{x}^{2}}>4\) or \(4{{x}^{2}}\le 28x\), or as complicated as \(2{{x}^{2}}7x\le 3\) or \(\displaystyle {{x}^{2}}+5x9<0\). We’ll use these as examples below.
Remember if you have a negative coefficient of \(\boldsymbol{{{x}^{2}}}\), you can move everything to the other side to make it positive – but be careful of the inequality signs!
There are three main methods used to solve Quadratic Inequalities. The Sign Pattern or Sign Chart Method is the most preferred, but I’ll cover a couple of methods here first.
Solving Using Graphing
You can solve quadratic inequalities by graphing the two sides of an inequality and seeing what the \(x\) intervals are for where one graph lies either below (\(<\)) or above (\(>\)) the other one.
Here are some examples using a Graphing Calculator.
What we’ll do is put in the left part of the equation in \({{Y}_{1}}=\) and the right part in \({{Y}_{2}}=\), see where they cross, and check which intervals are either greater than or less than, depending on the problem:
Graphing Calculator Instructions 
Screen 
\({{x}^{2}}>4\)
We can put \({{x}^{2}}\) in \({{Y}_{1}}=\) and 4 in \({{Y}_{2}}=\) and graph.
You can use the TRACE button (then arrows) to move closer to each intersection. Then can use the Intercept function to find the points of intersection: Push “2^{nd} TRACE” (CALC), and then either push 5, or move cursor down to intersect. You should see “First curve?” at the bottom. Then push ENTER, ENTER, ENTER.
Since we want the intervals where \({{x}^{2}}>4\), we need to see where the curved graph (parabola) is greater or higher than the line. We can see that this is \(\displaystyle \left( {\infty ,2} \right)\cup \left( {2,\infty } \right)\). (Remember that our answer will be in intervals along the \(x\)axis.) 

\(2{{x}^{2}}7x\le 3\)
Let’s do the same here by putting \(2{{x}^{2}}7x\) in \({{Y}_{1}}=\) and –3 in \({{Y}_{2}}=\), and graph.
You can use the TRACE button (then arrows) to move closer to each intersection before finding that intersection. Then can use the Intercept function to find the points of intersection: Push “2^{nd} TRACE” (CALC), and then either push 5, or move cursor down to intersect. You should see “First curve?” at the bottom. Then push ENTER, ENTER, ENTER.
Since we want the intervals where \(2{{x}^{2}}7x\le 3\), we need to see where the curved graph (parabola) is less than or lower than the line. We can see that this is \(\displaystyle \left[ {.5\,\,,3} \right]\). (Remember that our answer will be in intervals along the \(x\)axis.) 
Solving Algebraically, including Completing the Square
When solving algebraically, take the square root of each side, but we need to worry about the inequality sign. Break the equation into two equations like we did in the Solving Absolute Value Inequalities section (one with a plus, one with a minus), but the equation with the minus must have an inequality sign change.
For example, if we have \({{x}^{2}}<4\), we are saying the absolute value of \(x\) is less than the square root of 4, or \(\left x \right<2\). This is because the square root may be either positive or negative.
When you take the square root of both sides, divide up the equation into two equations. We get the first equation by just taking away the absolute value sign away on the left. The easiest way to get the second equation is to take the absolute value sign away on the left, and do two things on the right: reverse the inequality sign, and change the sign of everything on the right (even if we have variables over there).
Remember that with \(>\), we have an “or” and with \(<\) we have an “and”.
Here is an example:
Here is the example where we have to Complete the Square first:
Math 
Notes 
\(\displaystyle 2{{x}^{2}}7x\le 3\)
\(\displaystyle 2\left( {{{x}^{2}}\frac{7}{2}x+\,\,\underline{{\,\,\,\,\,\,\,\,}}\,} \right)\,\le \,3\,+\underline{{\,\,\,\,\,\,\,\,}}\) \(\displaystyle 2{{\left( {x\frac{7}{4}} \right)}^{2}}\le \,3+\underline{{\,\,\,\,\,\,\,\,}}\) 
Keep the –3 on the right handside and complete the square on the lefthand side. We don’t have to worry about the inequality sign until the end.
We must first divide through by whatever’s in front of the \({{x}^{2}}\), which is 2.
To complete the square, we need to now divide \(\displaystyle \frac{7}{2}\) by 2 to get \(\displaystyle \frac{7}{4}\), and then we’ll square it to get \(\displaystyle \frac{{49}}{{16}}\). We have to be careful though, since whatever we add to the lefthand side, we must add to the righthand side. 
\(\displaystyle \color{#2E8B57}{2}\left( {{{x}^{2}}\frac{7}{2}x+\,\,\underline{{\,\color{#2E8B57}{{\,\frac{{49}}{{16}}}}\,\,\,\,}}\,} \right)\le 3+\underline{{\color{#2E8B57}{{(2)\,\,\frac{{49}}{{16}}\,}}\,\,}}\)
\(\require{cancel} \displaystyle 2{{\left( {x\frac{7}{4}} \right)}^{2}}\le 3\,+\,\underline{{\,({}^{1}\cancel{2})\,\,\frac{{49}}{{{{{\cancel{{16}}}}_{8}}}}\,\,\,}}\) \(\displaystyle 2{{\left( {x\frac{7}{4}} \right)}^{2}}\le \,\,\frac{{24}}{8}\,+\frac{{49}}{8}\,\) \(\displaystyle 2{{\left( {x\frac{7}{4}} \right)}^{2}}\le \,\frac{{25}}{8}\,\) 
Note that since we added an extra 2 times \(\displaystyle \frac{{49}}{{16}}\) to the lefthand side, we need to also add it to the right.
Simplify (combine terms) on the righthand side. 
\(\displaystyle \,{{\left( {x\frac{7}{4}} \right)}^{2}}\le \frac{{25}}{{16}}\)
\(\displaystyle \sqrt{{\,\,{{{\left( {x\frac{7}{4}} \right)}}^{2}}\,}}\le \,\sqrt{{\frac{{25}}{{16}}}}\) \(\displaystyle \left {x\frac{7}{4}} \right\le \,\frac{5}{4}\)
\(\displaystyle x\frac{7}{4}\le \frac{5}{4}\text{ }\,\text{ }\,\text{and }\,\,x\frac{7}{4}\ge \frac{5}{4}\) \(\displaystyle x\le 3\text{ }\,\text{ }\,\text{and }\,\,\,\text{ }x\ge \frac{1}{2}\) \(\displaystyle \,\,\left[ {.5,\,\,3} \right]\) 
Divide by 2 to get the \(\displaystyle {{\left( {x\frac{7}{4}} \right)}^{2}}\) by itself. We now can take the square root of both sides.
Remember again that when you take the square root of both sides, you have to take the absolute value of the side where there is a variable. (This is because a square root can be either positive or negative.)
To solve, divide up the equation into two equations: the first by just taking away the absolute value sign, and the second by reversing the inequality sign, and changing the sign of everything on the right.
Solve both and put together in interval notation (we have an “and” because of the \(\le \)). Yeah! That’s what we got with the graphing calculator! 
Sign Chart (Sign Pattern) Method – the Easiest Method!
OK, so I’d love to introduce you to the sign chart or sign pattern method – a method that you’ll use later in Algebra when you work with Solving Polynomial Inequalities in the Graphing and Finding Roots of Polynomial Functions section and Rational Inequalities in the Graphing Rational Functions, including Asymptotes section. It looks difficult at first, but really isn’t too bad at all!
A sign chart or sign pattern is simply a number line that is separated into partitions (or intervals or regions), with boundary points (called “critical values“) that you get by setting the quadratic to 0 (without the inequality) and solving for \(x\) (the roots).
Sign charts are easy and a lot of fun since you can pick any point in between the critical values, and see if the whole quadratic is positive or negative. Then you just pick that interval (or intervals) by looking at the inequality.
Also, it’s a good idea to put open or closed circles on the critical values to remind ourselves if we have inclusive points (inequalities with equal signs, such as \(\le \) and \(\ge \)) or exclusive points (inequalities without equal signs, or factors in the denominators).
Let’s do some examples:
Sign Chart Problem 
Notes 
\(\begin{array}{c}\color{#800000}{{4{{x}^{2}}\le 28x}}\\\\4{{x}^{2}}28x\le 0\\4x\left( {x7} \right)\le 0\end{array}\) The problem calls for \(\le 0\), so we look for the minus sign(s), and our answers are inclusive (hard brackets).
The answer is \([0,7]\). 
The first thing we need to do is to get everything on the left side, and 0 on the right side, so we can see if we can factor the quadratic. Then we factor the quadratic (if we can). If we can’t, we’ll have to use the quadratic formula, or complete the square. For this quadratic, take out a Greatest Common Factor (GCF).
We now draw a sign chart. The boundary points, or critical values, are the roots (setting the factors to 0) of the quadratic, as if it were an equality.
Since the roots are 0 and 7 (we ignore the 4, since it’s a factor without a variable in it), we put those on the sign chart as boundaries. Then we check each interval with random points to see if the factored form of the quadratic is positive or negative. We put the signs over the interval.
I like to use the factored form for the check, since it’s a little easier for checking whether the random points are positive or negative. For example, we can try –1 for the leftmost interval: \(4\left( {1} \right)\left( {17} \right)=32\), which is \(+\). Let’s try 1 for the middle interval: \(4\left( 1 \right)\left( {17} \right)=24\), which is \(\). Let’s try 8 for the rightmost interval: \(4\left( 8 \right)\left( {87} \right)=32\), which is \(+\). Always try easy numbers, especially 0, if it’s not a critical value!
We want \(\le \) from the problem, so we look for the \(\) (negative) sign intervals; the interval is \([0,7]\). We need the hard brackets (inclusion of endpoints) since we have \(\le \) and not \(<\). See – it’s not that bad! 
Note: If there are no squares of any of the factors (with variables) in the quadratic factored form, the sign chart will typically be alternating minus and plus, like plusminusplus, or minusplusminus. If one of the factors is raised to an even factor (like squared), then this indicates a “bounce” in the graph, and the signs won’t change at that point.
Here are more examples:
Quadratic Inequality Problem 
Notes 
\(\begin{array}{c}\color{#800000}{{2{{x}^{2}}7x>3}}\\2{{x}^{2}}7x+3>0\\\,\left( {2x1} \right)\left( {x3} \right)>0\end{array}\) The problem calls for \(>0\), so we look for the plus sign(s), and our answers are exclusive (soft brackets). 
The first thing we need to do is to get everything on the lefthand side, and 0 on the righthand side, so we can see if we can factor the quadratic.
Since we can factor the quadratic, we’ll make a sign chart. We get the boundaries from getting the roots as if it were an equality; the critical values are \(\displaystyle \frac{1}{2}\) and 3 (set each factor to 0 and solve for \(x\)).
Then we check each interval with random points to see if the factored form of the quadratic is positive or negative. We put the signs over the interval. Make sure to use 0 as a test point if you can!
Look for the \(+\) intervals because of the \(>\). The interval is \(\displaystyle \left( {\infty ,\frac{1}{2}} \right)\cup \left( {3,\infty } \right)\). 
\({{\left( {x1} \right)}^{2}}\left( {3x+2} \right)<0\) The problem calls for \(<0\), so we look for the minus sign(s), and our answers are exclusive (soft brackets). 
The inequality is set up to make a sign chart, since we have 0 on the righthand side. The boundaries are \(\displaystyle \frac{2}{3}\) and 1.
Check each interval with random points (using “0”) and put the signs over the interval.
Note the “bounce” (no sign change) at \(x=1\) (for the \({{\left( {x1} \right)}^{2}}\)). This is because squaring an odd number still results in an even number.
Look for the \(\) intervals because of the \(<\). We have to be careful though, since we don’t include the boundary points (we have a \(<\) and not a \(\le \)). The interval then is \(\displaystyle \left( {\frac{2}{3},1} \right)\cup \left( {1,\infty } \right)\). 
And one more example, where we can’t factor:
Quadratic Inequality Problem 
Notes 
\(\displaystyle \begin{array}{c}\color{#800000}{{{{x}^{2}}+5x9\le 0}}\\\\\,x=\,\,\,\,\frac{{b\pm \sqrt{{{{b}^{2}}4ac}}}}{{2a}}\\x=\frac{{5\sqrt{{{{5}^{2}}4\left( 1 \right)\left( {9} \right)}}}}{2}\,\,\,\,\,\,\,\,\,\,x=\frac{{5+\sqrt{{{{5}^{2}}4\left( 1 \right)\left( {9} \right)}}}}{2}\\\,\,\,\,\,\,\,\,\,\,x=\frac{{5\sqrt{{61}}}}{2}\approx 6.41\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{{5+\sqrt{{61}}}}{2}\approx 1.41\end{array}\)
The problem calls for \(\le 0\), so we look for the minus sign(s), and our answers are inclusive (hard brackets). 
The first thing we need to do is to get everything on the lefthand side, and 0 on the righthand side, so we can see if we can factor the quadratic.
We can’t factor this one, so we’ll use the Quadratic Equation to find roots (as if it’s an equality). It might be easier just to use the Completing the Square method above, but we can do it this way, too.
The boundaries for the sign chart are these roots, which are approximately – 6.41 and 1.41.
Look for the \(\) intervals because of the \(\le \). The interval is \(\displaystyle \left[ {\frac{{5\sqrt{{61}}}}{2},\,\frac{{5+\sqrt{{61}}}}{2}} \right]\). The decimal (nonexact answer) is \(\left[ {6.41,\,1.41} \right]\). 
Real World Quadratic Inequality Example
Supposed Aven drops a ball off the top of a 10 foot pool slide, and the ball follows the projectile \(h\left( t \right)=16{{t}^{2}}+6\), where \(t\) is the time in seconds, and \(h\) is the height of the ball. Her friend Riley needs to catch the ball between 2 feet and 5 feet off the top of the water (ground). Between what two times should Riley try to catch the ball?
Solution:
Since we need to know when (\(t\)) the ball should be caught, we need to solve for \(t\), use an inequality since it can be anytime between 2 and 5 seconds. We’ll use just \(<\) signs since the problem says “between” and not “inclusive”. So let’s solve:
Solve Algebraically  Solve with Graphing Calculator 
\(2<16{{t}^{2}}+6<5\)
Let’s solve for \(t\): \(\displaystyle \begin{array}{l}2<16{{t}^{2}}+6\,\,\,\,\,\text{and}\,\,\,\,16{{t}^{2}}+6<5\\\,\,\,\,\,\,\,\,\,16{{t}^{2}}<4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,16{{t}^{2}}>1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{t}^{2}}<\frac{1}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{t}^{2}}>\frac{1}{{16}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t<\frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t>\frac{1}{4}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{4}<\,\,t<\frac{1}{2}\end{array}\)
When we take the square roots, we don’t need to worry about the negative values, since \(t\) can’t be negative.
Riley should try to catch the ball between \(\displaystyle \frac{1}{4}\) and \(\displaystyle \frac{1}{2}\) seconds. Good luck, Riley! 
I used 2^{nd} TRACE (CALC), 5 (intersect) to get the intersection of the quadratic and each of the lines. When the calculator asks for First Curve? and Second Curve?, just move the arrow buttons until it lands on the curves you want. Hit ENTER after Guess? to get the intersections.

Learn these rules, and practice, practice, practice!
Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.
If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software. You can even get math worksheets.
You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!
On to Quadratic Applications – you are ready!