Inequality with square root on both sides 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?

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 $\sqrt a<\sqrt b\iff a< b$
