Inequality with square root on both sides

Question

I have equation like this:

$$\sqrt{\vphantom{|}\ 3x^2-7x-20}<\sqrt{\vphantom{|}\ 8x+22}$$

I'm unsure how to solve it. I'm guessing I have to square both sides, but I don't know what happens with the inequality sign.

I guess there are four cases, depending on the sign of each side of the equation $(++,\ +-,\ -+,\ --)$.

And then I have to check if the solution fits the case, ie if both sides are of apropriate sign for the resulting interval.

But all this seems like a lot of work. There is an easier way, right?

How is it usually done?

Answer 1

"There is an easier way, right?"

Nope.

First things first: in order for these square root signs to even make sense, what's underneath them has to be positive. So, first order of business: work out for which values of $x$ both $3x^2-7x-20$ and $8x+22$ are positive.

Now, if $a$ and $b$ are positive numbers, then $\sqrt a < \sqrt b$ if and only if $a < b$. So once you've determined for which value of $x$ the inequality makes sense, you can just let $x$ be any such value, and just remove the square root signs. You'll end up with

$$3x^2-7x-20<8x+22$$

Now work out, out of the set of values of $x$ you found in the first part of the problem, which ones verify that inequality.

Answer 2

HINT:

As $3x^2-7x-20=(3x+5)(x-4)$

$\sqrt{3x^2-7x-20}$ will be real if $x\ge$max$(4,-5/3)=4$ or if $x\le$min$(4,-5/3)=-5/3$

Now $\sqrt{8x+22}$ will be real if $8x+22\ge0\iff x\ge-11/4$

So, we need $-11/4\le x\le-5/3$ and for $x\ge4$

Now use $\sqrt a<\sqrt b\iff a< b$