Solve the inequality $\sqrt{(x+2)(x-5)} < 8-x$

Question

Solve the inequality $\sqrt{(x+2)(x-5)} < 8-x$.

I got the correct solution, but I want to make sure I'm doing everything properly.

First I wrote

$(x+2)(x-5)\geq 0$ $\iff (x+2\geq 0 \land x-5\geq 0) \lor (x+2\leq 0 \land x-5\leq 0) \iff x\geq 5 \lor x\leq -2 \iff x\in(-\infty,-2]\cup[5, \infty)$

$(x+2)(x-5)<64-16x+x^{2}$

$x^{2}-3x-10<64-16x+x^{2}$

$x<\frac{74}{13}$

$x\in(-\infty,-2]\cup[5,\frac{74}{13})$

Is the usage of $\iff$ correct? Should I put $\iff$ somewhere else also? Am I even allowed to square this inequality?

Answer 1

The domain of this inequation is given by the condition $(x+2)(x-5)\ge0$, i.e., by the theorem on the sign of quadatic polynomials, $(-\infty,-2]\cup[5,+\infty)$.

Remember that, on domain of the inequation $\sqrt A< B \iff \bigl((A<B^2)\quad\textbf{and}\quad B\ge0$.

Now $\;(x+2)(x-5)<(8-x)^2\iff 13x <74$, so the condition $x<8$ is automatically satisfied, and finally, the set of solutions is $$\bigl(-\infty,-2\bigr]\cup\bigl[5,\tfrac{74}{13}\bigr).$$

Answer 2

Looks like your solution is correct, as is your use of $\iff$.

Yes, you are allowed to square the inequality because inequalities hold under squaring if both sides are nonnegative.