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Wize High School Algebra II Textbook (Common Core) > Quadratic Equations & Complex Numbers

Solving Quadratic Inequalities in One Variable

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Solving Quadratic Inequalities in One Variable

A quadratic inequality in one variable looks something like this: ax2+bx+c>0ax^2+bx+c >0.
In this case, we are asking: "for what values of xx is the parabola above the x-axis (positive)?"
*The inequality can be any one of >, ≥, <, ≤>,\ \ge,\ <,\ \le.

Step 1.
Identify the x-intercepts ⟶\longrightarrow rearrange if necessary, then factor OR use the quadratic formula.

Step 2.
Choose test points to the left and right of each x-intercept.

Step 3.
Create a table and a number line. Use the test points to see if the function is positive or negative in that region.

Step 4.
Express the answer using interval notation or set notation.

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Example
Let's look at the polynomial inequality (x−2)(x+4)≥0(x-2)(x+4)\geq{}0.

Step 1.
It is already in factored form.
xx-intercepts: x=−4,2x=-4,2

Step 2.
Choose test points around the x-intercepts: x=−5,0,3x=-5,0,3

Step 3.
Create a table with rows containing: the intervals around the x-intercepts, the chosen test points, and the quadratic we are checking.
Fill in the last row by determining whether the quadratic (x−2)(x+4)(x-2)(x+4) is positive or negative when the test point is plugged in.
E.g. Using test point x=−5 ⟶  (−5−2)(−5+4)=7 ⟶+x=\colorFour{-5}\ \longrightarrow\ \ (\colorFour{-5}-2)(\colorFour{-5}+4)=7\ \longrightarrow \colorbox{yellow}{$+$}

Intervalx<−4−4<x<2x>2Test Point−503(x−2)(x+4)+−+\begin{array}{|c|c|c|c|} \hline\\ \text{Interval}& x<-4 & -4< x<2 & x>2 \\\\ \hline\\ \text{Test Point}&-5&0&3\\\\ \hline\\ (x-2)(x+4)&\bm{\colorbox{yellow}{$+$}} &\bm{\colorbox{aqua}{$-$}} &\bm{\colorbox{yellow}{$+$}}\\\\\hline \end{array}

Step 4.
The solution depends on the inequality symbol. We are looking at (x−2)(x+4)≥0(x-2)(x+4) \colorbox{yellow}{$\ge0$} .

We want the part that lies above (or is equal to) 0, so we want the intervals where it is positive (+\colorbox{yellow}{$+$}).

The solution is therefore x≤−4x\le-4 or x≥2x\ge2.
In interval notation, we have (−∞,−4] ∪ [2,∞)(-\infin,-4\bm]~\cup~\bm[2,\infin).
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Alternative Method

Instead, we can simply graph the quadratic.
If we know the x-intercepts, we can easily tell exactly where the graph is positive or negative!


We know that the regions are separated by the x-intercepts.
We can see that (x−2)(x+4)(x-2)(x+4) is positive on the left side and the right side, but not in the center. Therefore:

{x∈R ∣ x≤−4  or  x≥2}\{ x \in \reals \ |\ x\le-4 \ \text{ or }\ x \ge2\}
OR
(−∞,−4] ∪ [2,∞)(-\infin,-4\bm]~\cup~\bm[2,\infin)
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Example: Solving Quadratic Inequalities in One Variable

Solve −2x2>−x−6-2x^2>-x-6.

Step 1: Find x-intercepts

First, we have to rearrange so that everything is on one side. Subtracting xx and 66 from both sides, we get:

−2x2+x+6>0-2x^2+x+6>0

This cannot be easily factored, so we use the quadratic formula to find the x-intercepts:

x=−1±12−4(−2)(6)2(−2)=−1±49−4=−1±7−4  ⟹  x=−3/2, 2\begin{aligned} x &=\dfrac{-1\pm\sqrt{1^2-4(-2)(6)}}{2(-2)}\\[1em] &=\dfrac{-1\pm\sqrt{49}}{-4}\\[1em] &=\dfrac{-1\pm7}{-4}\\[1.5em] \implies x&=-3/2,\ 2 \end{aligned}

Step 2: Choose test points
  • For x<−3/2x<-3/2, we can choose x=−2x=-2
  • For −3/2<x<2-3/2<x<2, we can choose x=0x=0
  • For x>2x>2, we can choose x=3x=3
Step 3: Create a table

Intervalx<−3/2−4<x<2x>2Test Point−203−2x2+x+6−+−\begin{array}{|c|c|c|c|} \hline\\ \text{Interval}& x<-3/2 & -4<x<2 & x>2 \\\\ \hline\\ \text{Test Point}&-2&0&3\\\\\hline\\ -2x^2+x+6&\colorbox{aqua}{$-$} &\colorbox{yellow}{$+$} &\colorbox{aqua}{$-$}\\\\\hline \end{array}

Step 4: Interpret

Since we want to know when −2x2+x+6>0-2x^2+x+6\colorbox{yellow}{$>0$}, look for region(s) where it is positive (+\colorbox{yellow}{$+$}).

The solution is therefore: {x∈R ∣ −4<x<2}\{x\in \reals \ |\ -4<x<2\}
OR
In interval notation: (−4,2)(-4,2), where round parentheses are used instead of brackets since ">>" means "strictly greater than".

Practice: Solving Quadratic Inequalities in One Variable

Given the graph of the quadratic f(x)=x2−x−6f(x)=x^2-x-6, match the inequality with the correct solution.

A.
(−2,3)(-2,3)
B.
(−∞,−2) ∪ (3,∞)(-\infin,-2)~\cup~(3,\infin)
C.
(−∞,−2] ∪ [3,∞)(-\infin,-2\bm]~\cup~\bm[3,\infin)
D.
[−2,3]\bm[-2,3\bm]
x2−x−6<0x^2-x-6<0
x2−x−6≤0x^2-x-6\le0
x2−x−6>0x^2-x-6>0
x2−x−6≥0x^2-x-6\ge0

Practice: Solving Quadratic Inequalities in One Variable

Solve 3x2≤−9x3x^2\le-9x.

Practice: Solving Quadratic Inequalities in One Variable


The height of an isosceles is given as four times the length of its base.

A square has sides that are 1 unit longer than the base of the isosceles triangle.

What values for the length of the base of the triangle will ensure that the area of the triangle is larger than the area of the square?