What is a quadratic inequality?
A quadratic inequality has \(x^2\) and an inequality sign.
Examples:
\(x^2 - 5x + 6 \gt 0\)
\(x^2 - 4 \le 0\)
\(2x^2 + 3x - 2 \ge 0\)
A quadratic inequality asks:
For which values of \(x\) is the quadratic expression positive, negative, greater than zero, or less than zero?
The graph is a parabola.
Change the sign and values. The blue shadow shows the solution intervals.
The main strategy:
- Move everything to one side.
- Set the expression equal to zero.
- Solve the quadratic equation.
- Use the roots to divide the number line into intervals.
- Test each interval.
- Choose the interval that satisfies the inequality.
Key idea: roots split the number line
Suppose:
\((x - 2)(x - 3) \gt 0\)
The roots are:
\(x = 2,\quad x = 3\)
These divide the number line into three intervals:
\((-\infty, 2),\quad (2, 3),\quad (3, \infty)\)
Then we test each interval.
Quadratic Inequality Example 1
Solve:
\(x^2 - 5x + 6 \gt 0\)
Step 1: Factor.
\(x^2 - 5x + 6 = (x - 2)(x - 3)\)
So:
\((x - 2)(x - 3) \gt 0\)
Step 2: Find roots.
\(x - 2 = 0 \Rightarrow x = 2\)
\(x - 3 = 0 \Rightarrow x = 3\)
Step 3: Split number line.
\((-\infty, 2),\quad (2, 3),\quad (3, \infty)\)
Step 4: Test each interval.
Interval 1: \(x \lt 2\)
Choose \(x = 0\).
\((0 - 2)(0 - 3) = (-2)(-3) = 6\)
Positive. So this interval works.
Interval 2: \(2 \lt x \lt 3\)
Choose \(x = 2.5\).
\((2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25\)
Negative. Does not work.
Interval 3: \(x \gt 3\)
Choose \(x = 4\).
\((4 - 2)(4 - 3) = (2)(1) = 2\)
Positive. Works.
Answer:
\(x \lt 2\text{ or }x \gt 3\)
Interval notation:
\((-\infty, 2) \cup (3, \infty)\)
Important: because the inequality is \(\gt 0\), not \(\ge 0\), we do not include 2 and 3.
Quadratic Inequality Example 2
Solve:
\(x^2 - 4 \le 0\)
Step 1: Factor.
\(x^2 - 4 = (x - 2)(x + 2)\)
So:
\((x - 2)(x + 2) \le 0\)
Step 2: Find roots.
\(x - 2 = 0 \Rightarrow x = 2\)
\(x + 2 = 0 \Rightarrow x = -2\)
Step 3: Split number line.
\((-\infty, -2),\quad (-2, 2),\quad (2, \infty)\)
Step 4: Test each interval.
Interval 1: \(x \lt -2\)
Choose \(x = -3\).
\((-3)^2 - 4 = 9 - 4 = 5\)
Positive. We need \(\le 0\), so it does not work.
Interval 2: \(-2 \lt x \lt 2\)
Choose \(x = 0\).
\(0^2 - 4 = -4\)
Negative. This works.
Interval 3: \(x \gt 2\)
Choose \(x = 3\).
\(3^2 - 4 = 9 - 4 = 5\)
Positive. Does not work.
Step 5: Include endpoints?
The inequality is:
\(\le 0\)
So yes, include the roots \(-2\) and \(2\), because at those points the expression equals zero.
Answer:
\(-2 \le x \le 2\)
Interval notation:
\([-2, 2]\)
Quadratic Inequality Example 3
Solve:
\(2x^2 + 3x - 2 \ge 0\)
Step 1: Factor.
\(2x^2 + 3x - 2 = (2x - 1)(x + 2)\)
So:
\((2x - 1)(x + 2) \ge 0\)
Step 2: Find roots.
\(2x - 1 = 0\)
\(2x = 1\)
\(x = \dfrac{1}{2}\)
and:
\(x + 2 = 0\)
\(x = -2\)
Step 3: Split number line.
\((-\infty, -2),\quad \left(-2, \dfrac{1}{2}\right),\quad \left(\dfrac{1}{2}, \infty\right)\)
Step 4: Test intervals.
Interval 1: \(x \lt -2\)
Choose \(x = -3\).
\((2(-3) - 1)(-3 + 2) = (-7)(-1) = 7\)
Positive. Works.
Interval 2: \(-2 \lt x \lt \dfrac{1}{2}\)
Choose \(x = 0\).
\((2(0) - 1)(0 + 2) = (-1)(2) = -2\)
Negative. Does not work.
Interval 3: \(x \gt \dfrac{1}{2}\)
Choose \(x = 1\).
\((2(1) - 1)(1 + 2) = (1)(3) = 3\)
Positive. Works.
Step 5: Include endpoints?
The inequality is:
\(\ge 0\)
So yes, include the roots.
Answer:
\(x \le -2\text{ or }x \ge \dfrac{1}{2}\)
Interval notation:
\((-\infty, -2] \cup \left[\dfrac{1}{2}, \infty\right)\)