the square root of x+120 minus the square root of x+3 is equal to 9.

I got confused on simplifying part because I didnt get it to match 5 which is the answer.

If you can show me how to solve this the proper way please and how to post problems in here with square root signs and divsions and fractions please

Thank you!

Originally Posted by daword177
the square root of x+120 minus the square root of x+3 is equal to 9.
If you can show me how to solve this the proper way please and how to post problems in here with square root signs and divsions and fractions please.
Use LaTeX code: [TEX]\sqrt{x+120}-\sqrt{x+3}=9[/TEX] gives $\sqrt{x+120}-\sqrt{x+3}=9$

Add to both sides $\sqrt{x+120}=9+\sqrt{x+3}$
Square both sides ${x+120}=81+18\sqrt{x+3}+(x+3)$.
Combine terms ${36}=18\sqrt{x+3}$
Divide to get $\sqrt{x+3}=2$
Finish.

Note that x is NOT equal to 5. If x= 5, then $\sqrt{x+ 120}= \sqrt{5+ 120}= \sqrt{125}= 5\sqrt{5}$ and $\sqrt{x+ 3}= \sqrt{5+ 3}= \sqrt{8}= 2\sqrt{2}$. $\sqrt{x+120}- \sqrt{x+ 3}= 5\sqrt{5}- 2\sqrt{2}\ne 9$.