Question

Solve the inequality √(x²-2x-8) ≤ -x+2

Answer 1

$\sqrt {x^2-2x-8}\leq -x+2$ $...(1)$

We know that, if $\sqrt {f(x)}\leq g(x)$ then $g(x) \geq0$

So, $-x+2\geq0\Rightarrow x\leq2 \ ...(2)$

As $-x+2\geq0$

So, squaring both sides in equation (1), we get

$x^2-2x-8\leq(-x+2)^2 \\\Rightarrow x^2-2x-8\leq x^2+4-4x \\\Rightarrow 2x\leq12 \\\Rightarrow x\leq6 \ ...(3)$

Also $x^2-2x-8\geq 0$ as value inside square root can not be negative.

$\\\Rightarrow (x+2)(x-4)\geq0$

From the above, we can say that

$x\leq-2$ or $x\geq4 \ ...(4)$

From equation (2), (3) and (4)

From the graph, we can say that the common solution is

$x\leq-2$