Quadratic Inequalities: Practice

The roots split the number line into three intervals.

$(x - 1)(x - 4) = 0 \Rightarrow x = 1 \text{ or } x = 4$

The upward parabola is positive outside the roots, and the strict $> 0$ excludes the endpoints, so the solution is $x < 1$ or $x > 4$.

Find the roots first.

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

The upward parabola is negative between the roots, and the inclusive $\le 0$ keeps the endpoints, so the solution is $-2 \le x \le 3$.

The roots are where the product equals zero.

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

The upward parabola is positive outside the roots, and the inclusive $\ge 0$ keeps the endpoints, so the solution is $x \le -2$ or $x \ge 3$.

Factor the difference of squares.

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

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

Use the same roots as the strict version.

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

The upward parabola is negative between the roots, and the inclusive $\le 0$ keeps the endpoints, so $-2 \le x \le 2$.

The roots are where each factor is zero.

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

The upward parabola is positive outside the roots, and the inclusive $\ge 0$ keeps them, so $x \le -1$ or $x \ge 5$.

The curve meets the axis at its roots.

$(x - 2)(x - 6) = 0 \Rightarrow x = 2 \text{ or } x = 6$

An upward parabola is above the axis (so $y > 0$) outside the roots, and 'above' is strict, so the solution is $x < 2$ or $x > 6$.

Factor out the common $x$.

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

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

Factor the difference of squares.

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

The upward parabola is negative between the roots, and the strict symbol excludes them, so $-3 < x < 3$.

Factor out the common $x$.

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

The upward parabola is negative between the roots, and the strict symbol excludes them, so $-3 < x < 0$.

The roots are where each factor is zero.

$(x + 4)(x + 1) = 0 \Rightarrow x = -4 \text{ or } x = -1$

The upward parabola is positive outside the roots, and the strict symbol excludes them, so $x < -4$ or $x > -1$.

The roots split the line into three pieces.

$(x + 4)(x + 1) = 0 \Rightarrow x = -4 \text{ or } x = -1$

The upward parabola is negative between the roots, and the strict symbol excludes them, so $-4 < x < -1$.