Quadratic Inequalities: Practice

Factor to find the roots.

$x^2 - 7x + 12 = (x - 3)(x - 4) = 0 \Rightarrow x = 3 \text{ or } x = 4$

The upward parabola is negative between the roots, and the inclusive $\le 0$ keeps the endpoints, so $3 \le x \le 4$.

Factor to find the roots.

$x^2 + 2x - 8 = (x + 4)(x - 2) = 0 \Rightarrow x = -4 \text{ or } x = 2$

The upward parabola is positive outside the roots, and the strict symbol excludes them, so $x < -4$ or $x > 2$.

Multiply both sides by $-1$ and flip the symbol to face an upward parabola.

$-x^2 + x + 6 \ge 0 \Rightarrow x^2 - x - 6 = (x - 3)(x + 2) \le 0$

The roots are $-2$ and $3$; the upward parabola is negative between them, and the inclusive symbol keeps the endpoints, so $-2 \le x \le 3$.

Rearrange to compare a square with a number.

$-x^2 + 4 > 0 \Rightarrow x^2 < 4 \Rightarrow (x - 2)(x + 2) < 0$

The roots are $-2$ and $2$; the product is negative between them, so $-2 < x < 2$.

Factor the quadratic with a leading coefficient of $2$.

$2x^2 - 5x - 3 = (2x + 1)(x - 3) = 0 \Rightarrow x = -\tfrac{1}{2} \text{ or } x = 3$

The upward parabola is negative between the roots, and the inclusive symbol keeps the endpoints, so $-\tfrac{1}{2} \le x \le 3$.

Move every term to one side, then factor.

$x^2 \ge 5x \Rightarrow x^2 - 5x = x(x - 5) \ge 0 \Rightarrow x = 0 \text{ or } x = 5$

The upward parabola is positive outside the roots, and the inclusive symbol keeps them, so $x \le 0$ or $x \ge 5$.

Bring every term to the left, then factor.

$x^2 < 3x + 10 \Rightarrow x^2 - 3x - 10 = (x - 5)(x + 2) < 0 \Rightarrow x = -2 \text{ or } x = 5$

The product is negative between the roots, so $-2 < x < 5$.

Factor to find the roots.

$x^2 + x - 12 = (x + 4)(x - 3) = 0 \Rightarrow x = -4 \text{ or } x = 3$

The upward parabola is negative between the roots, and the inclusive $\le 0$ keeps the endpoints, so $-4 \le x \le 3$.

Move everything to one side and factor.

$x^2 > 9 \Rightarrow x^2 - 9 = (x - 3)(x + 3) > 0 \Rightarrow x = -3 \text{ or } x = 3$

The upward parabola is positive outside the roots, and the strict symbol excludes them, so $x < -3$ or $x > 3$.

Move everything to one side and factor.

$x^2 \le 1 \Rightarrow x^2 - 1 = (x - 1)(x + 1) \le 0 \Rightarrow x = -1 \text{ or } x = 1$

The upward parabola is negative between the roots, and the inclusive $\le 0$ keeps the endpoints, so $-1 \le x \le 1$.

Factor to find the roots.

$x^2 + 6x + 5 = (x + 1)(x + 5) = 0 \Rightarrow x = -5 \text{ or } x = -1$

The upward parabola is negative between the roots, and the strict symbol excludes them, so $-5 < x < -1$.

Factor out $x$; the roots are $0$ and $4$.

$4x - x^2 = x(4 - x) = 0 \Rightarrow x = 0 \text{ or } x = 4$

Between the roots the product $x(4 - x)$ is positive, and the inclusive $\ge 0$ keeps the endpoints, so $0 \le x \le 4$.