How do you solve $x^{2} - 2 x \leq 8$ $x^2 -2x ≤ 8$ ?

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Answer 1 · 2015-08-30T16:50:27

$- 2 \leq x \leq 4$ $-2<=x<=4$

Factorise the expression:

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

i) $x - 4 \geq 0 and x + 2 \leq 0$ $x-4>=0 and x+2<=0$ (ii) $x - 4 \leq 0 and x + 2 \geq 0$ $x-4<=0 and x+2>=0$

In the i) case , $x \geq 4 and x \leq - 2$ $x>=4 and x<=-2$ , which is not possible.

In ii) case $x \leq 4 and x \geq - 2$ $x<=4 and x>=-2$ is a feasible solution. The solution is therefore $- 2 \leq x \leq 4$ $-2<=x<=4$

How do you solve x2−2x≤8x^2 -2x ≤ 8?