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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?

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    $\begingroup$ Your forgot the condition $x\le 8$ to have an equivalence when squaring (per chance, it is satisfied by the solutions you found). As to the sign of $(x+2)(x-5)$, there is a theorem on the sign of a quadratic polynomial, so you can shorten your argument. $\endgroup$ – Bernard Nov 1 '16 at 19:20
  • $\begingroup$ @Bernard So I should have 2 cases, one where $x\leq 8$ where I get the solution I already found, and the other when $x>8$ where I should switch the inequality sign when squaring my inequality. In that case I get $x\in (8,\infty)$. Is that right? $\endgroup$ – lmc Nov 1 '16 at 19:30
  • $\begingroup$ No. What I mean is that $\sqrt A <B\iff (A<B^2\quad\textbf{and}\quad B\ge 0)$ — and $A\ge 0$, of course. $\endgroup$ – Bernard Nov 1 '16 at 19:32
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    $\begingroup$ No. If $x>8$, LHS is positive, and can't be less than $0$. So there are no solutions in this case. What Bernard said is that $a^2<b^2$ is equivalent to $a<b$ iff $0\le a,b$. $\endgroup$ – Nicolas FRANCOIS Nov 1 '16 at 19:34
  • $\begingroup$ Okay. And just to make sure, $\sqrt A>B \iff A>B^{2}$ and $B\geq 0$? $\endgroup$ – lmc Nov 1 '16 at 19:37
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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 $$\bigl(-\infty,-2\bigr]\cup\bigl[5,\tfrac{74}{13}\bigr).$$

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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.

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