Unit 9 · Lesson 5 of 5
Quadratic inequalities
You'll be able to: solve inequalities like \(x^2 - x - 6 \lt 0\) and write the answer in interval notation.
Step 1 of 5Warm-up
Two quick questions from earlier lessons. Answer from memory first. Today joins these two skills together.
Find the vertex of \(y = x^2 - 2x - 3\).
\(x = -\dfrac{-2}{2(1)} = 1\). Then \(y = 1 - 2 - 3 = -4\).
Answer: \((1, -4)\).
Solve \(-3x + 5 \le 11\). Write the answer in interval notation.
Subtract 5: \(-3x \le 6\). Divide by \(-3\) and flip the sign: \(x \ge -2\).
Answer: \([-2, \infty)\).
Step 2 of 5The idea
A quadratic inequality asks where an expression like \(x^2 - x - 6\) is negative (\(\lt 0\)) or positive (\(\gt 0\)). The expression can only switch from positive to negative where it equals 0. So those zeros cut the number line into pieces, and each piece is all positive or all negative.
Steps.
1. Move everything to one side, so the other side is 0.
2. Solve the equation \(= 0\) (factor, or use the quadratic formula). These are the boundary points.
3. Mark them on a number line. Test one number from each piece in the inequality.
4. Keep the pieces that make it true. Use [ ] at the boundary points for \(\le\) or \(\ge\), and ( ) for \(\lt\) or \(\gt\).
The graph gives a shortcut. If \(a \gt 0\), the parabola opens up: it is below the x-axis (\(\lt 0\)) between the zeros and above it (\(\gt 0\)) outside them. An "outside" answer is two pieces joined with \(\cup\), a union.
Write the answer in interval notation. Here you can type U for \(\cup\), and inf for \(\infty\).
Why this shows up on the placement test: it checks factoring, solving, and interval notation in one question, so it is a common test item.
Step 3 of 5Worked examples
Example 1: \(x^2 - x - 6 \lt 0\)
- It is already "something \(\lt 0\)". Solve \(x^2 - x - 6 = 0\): \((x - 3)(x + 2) = 0\), so \(x = 3\) or \(x = -2\).
- The zeros cut the line into three pieces: left of \(-2\), between \(-2\) and \(3\), right of \(3\).
- Test \(x = -3\): \(9 + 3 - 6 = 6\). Is \(6 \lt 0\)? No.
- Test \(x = 0\): \(0 - 0 - 6 = -6\). Is \(-6 \lt 0\)? Yes.
- Test \(x = 4\): \(16 - 4 - 6 = 6\). No.
- Only the middle piece works, and \(\lt\) leaves the zeros out. Answer: \((-2, 3)\). The graph agrees: it dips below the x-axis only between \(-2\) and \(3\).
Example 2: \(x^2 + 4x \ge 5\)
- Get 0 on one side: \(x^2 + 4x - 5 \ge 0\). Testing only works against 0.
- Zeros: \((x + 5)(x - 1) = 0\), so \(x = -5\) or \(x = 1\).
- Test \(x = -6\): \(36 - 24 - 5 = 7 \ge 0\). Yes.
- Test \(x = 0\): \(-5 \ge 0\). No.
- Test \(x = 2\): \(4 + 8 - 5 = 7 \ge 0\). Yes.
- The two outside pieces work, and \(\ge\) includes the zeros. Answer: \((-\infty, -5] \cup [1, \infty)\).
Step 4 of 5Practice
Type your answer and press Check. Use interval notation like (-inf, 2) U [5, inf), or an inequality like x < 2 or x >= 5. If you miss twice, open the worked solution. Then mark how it went.
Solve \((x - 1)(x + 4) \gt 0\).
Zeros: \(x = 1\) and \(x = -4\).
Test \(x = -5\): \((-6)(-1) = 6 \gt 0\), yes. Test \(x = 0\): \((-1)(4) = -4\), no. Test \(x = 2\): \((1)(6) = 6\), yes.
Answer: \((-\infty, -4) \cup (1, \infty)\).
Solve \(x^2 - 9 \le 0\).
Zeros: \(x^2 = 9\), so \(x = \pm 3\).
Test \(x = 0\): \(-9 \le 0\), yes. Test \(x = \pm 4\): \(16 - 9 = 7\), no.
\(\le\) includes the zeros. Answer: \([-3, 3]\).
Solve \(x^2 + 2x - 15 \lt 0\).
Zeros: \((x + 5)(x - 3) = 0\), so \(x = -5\) or \(x = 3\).
Test \(x = 0\): \(-15 \lt 0\), yes. Test \(x = -6\): \(36 - 12 - 15 = 9\), no. Test \(x = 4\): \(16 + 8 - 15 = 9\), no.
Answer: \((-5, 3)\).
Solve \(x^2 \ge 4x\).
Get 0 on one side: \(x^2 - 4x \ge 0\). Factor: \(x(x - 4) = 0\), so \(x = 0\) or \(x = 4\). Don't divide by \(x\): you would lose a zero.
Test \(x = -1\): \(1 + 4 = 5 \ge 0\), yes. Test \(x = 1\): \(1 - 4 = -3\), no. Test \(x = 5\): \(25 - 20 = 5\), yes.
Answer: \((-\infty, 0] \cup [4, \infty)\).
Solve \(2x^2 + 5x - 3 \le 0\).
Factor: \((2x - 1)(x + 3) = 0\), so \(x = \dfrac{1}{2}\) or \(x = -3\).
Test \(x = 0\): \(-3 \le 0\), yes. Test \(x = -4\): \(32 - 20 - 3 = 9\), no. Test \(x = 1\): \(2 + 5 - 3 = 4\), no.
Answer: \(\left[-3, \dfrac{1}{2}\right]\).
Solve \(x^2 + 2x + 5 \lt 0\).
Look for zeros: the discriminant is \(4 - 4(1)(5) = -16\), negative. So there are no real zeros, and the parabola never crosses the x-axis.
Test any number, say \(x = 0\): \(5 \lt 0\) is false. So it is false everywhere (the parabola opens up and stays above the axis).
Answer: no solution (none).
Solve \(x^2 - 2x - 1 \gt 0\).
This does not factor. Quadratic formula: \(x = \dfrac{2 \pm \sqrt{4 + 4}}{2} = \dfrac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}\). That is about \(-0.41\) and \(2.41\).
Test \(x = 0\) (between them): \(-1 \gt 0\), no. So the outside pieces work (opens up: above the axis outside the zeros).
Answer: \((-\infty, 1 - \sqrt{2}) \cup (1 + \sqrt{2}, \infty)\).
A ball's height in feet after \(t\) seconds is \(h = -16t^2 + 64t\). For which times \(t\) is the ball higher than 48 feet?
Solve \(-16t^2 + 64t \gt 48\). Get 0 on one side: \(-16t^2 + 64t - 48 \gt 0\).
Divide by \(-16\) and flip the sign: \(t^2 - 4t + 3 \lt 0\). Zeros: \((t - 1)(t - 3) = 0\), so \(t = 1\) or \(t = 3\).
Test \(t = 2\): \(4 - 8 + 3 = -1 \lt 0\), yes; outside pieces fail. Answer: \((1, 3)\), between 1 and 3 seconds.
Step 5 of 5Wrap-up
You can now solve a quadratic inequality: get 0 on one side, find the zeros, test a number in each piece, and write the pieces that work in interval notation.
Read or watch next: OpenStax Intermediate Algebra 2e, 9.8 Solve Quadratic Inequalities (free textbook, with more practice at the end of the section).
Also good: Khan Academy: Quadratic inequalities (short videos). Quick rules: Quadratics cheat sheet.
Something unclear? Ask your teacher bot a follow-up question, for example: "How can I tell from the graph whether the answer is between the zeros or outside them?" It knows this course and can explain things another way.