# How do you solve x^2 -2x ≤ 8?

Aug 30, 2015

$- 2 \le x \le 4$

#### Explanation:

Factorise the expression:

${x}^{2} - 2 x - 8 \le 0$
(x-4)(x+2)$\le$0
For this inequality to be true following two possibilities can exist:

i) $x - 4 \ge 0 \mathmr{and} x + 2 \le 0$ (ii)$x - 4 \le 0 \mathmr{and} x + 2 \ge 0$

In the i) case , $x \ge 4 \mathmr{and} x \le - 2$, which is not possible.

In ii) case$x \le 4 \mathmr{and} x \ge - 2$ is a feasible solution. The solution is therefore $- 2 \le x \le 4$