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?

  • $\begingroup$ Inequality isn't showing properly for me so here it is: sqrt(3x^2-7x-20)<sqrt(8x+22) $\endgroup$ – Karlovsky120 Mar 13 '17 at 11:54
  • $\begingroup$ There is only one side in your problem statement? Did you forget to type the other side? $\endgroup$ – Student Mar 13 '17 at 11:55
  • $\begingroup$ Are you sure you are not dealing only with positive square root? Because if $8x+22$ is negative (for example, $x= 3$), then we get a complex number as an answer, and there is no ordering among complex numbers. $\endgroup$ – Teresa Lisbon Mar 13 '17 at 11:57
  • $\begingroup$ I'm not sure. This is a problem regarding quadratic functions and inequalities. P.S. 8*3+22=46>0, I'm guessing you meant -3. $\endgroup$ – Karlovsky120 Mar 13 '17 at 11:59

"There is an easier way, right?"


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


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.

  • $\begingroup$ Thanks! Funny note; I checked for the solution in the back of the book, there was an explanation of the principle there. $\endgroup$ – Karlovsky120 Mar 13 '17 at 12:18


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


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