Coaching Centre
Quadratic equations and inequations
Solve quadratic inequations by sketching the parabola or using a table of test points, and write the answer as inequalities or intervals.
The graph method
The green band marks where y > 0. A concave up parabola is positive outside its zeroes; a concave down one is positive between them.
Solve $x^2 - x - 6 \le 0$.
Solve $2x^2 > 5x + 3$.
The table of test points
The sign of a quadratic can only change at a zero. So pick one test value in each region between the zeroes and check the sign.
| $x$ | $-3$ | $-2$ | $0$ | $3$ | $4$ |
|---|---|---|---|---|---|
| $x^2 - x - 6$ | $6$ | $0$ | $-6$ | $0$ | $6$ |
| sign | $+$ | zero | $-$ | zero | $+$ |
Solve $x^2 - 9 > 0$.
Concave up with zeroes $\pm 3$: positive outside them.
Solve $-x^2 + 4x \ge 0$.
$-x(x - 4) \ge 0$: concave down, so non-negative between the zeroes, including them.
$x^2 + 1 > 0$ is true for:
$x^2 \ge 0$, so $x^2 + 1 \ge 1 > 0$ always.
Solve $x^2 - 5x + 6 < 0$.
$(x - 2)(x - 3) < 0$: concave up, negative between the zeroes.
Solve $x^2 + 2x - 8 > 0$.
$(x + 4)(x - 2) > 0$: positive outside the zeroes.
Solve $x^2 \ge 16$.
$x^2 - 16 \ge 0$: outside the zeroes $\pm 4$, including them.
What is the smallest integer $x$ with $x^2 - 3x - 10 < 0$?
$(x - 5)(x + 2) < 0$ gives $-2 < x < 5$, so the smallest integer is $-1$.
$x^2 - 4x + 4 > 0$ is true for:
$(x - 2)^2 > 0$ everywhere except at $x = 2$, where it equals $0$.
New numbers every time. Keep going until you get five in a row.