# Thread: algebra problem I'm stuck on

1. ## algebra problem I'm stuck on

$\sqrt{\sqrt{x+5}+x}=5$

$\sqrt{x+5} + x = 25$

$(\sqrt{x+5})^2 = (25-x)^2$

$x+5 = 625 - 50x - x^2$

$0 = \frac{-x^2 = 49x + 620}{-1}$

$0 = x^2 + 49x - 620$

Plug that into the quadratic equation and I get:

x = $\frac{-49 \pm \sqrt{4881}}{2}$

Which I can't find a way to simplify down.

The book shows the answer to be 21. How they got it, I have no idea. So any help will be greatly appreciated

2. Originally Posted by mathgeek777
$\sqrt{\sqrt{x+5}+x}=5$

$\sqrt{x+5} + x = 25$

$(\sqrt{x+5})^2 = (25-x)^2$

$x+5 = 625 - 50x {\color{red}+} x^2$

...
From here, you should get $x^2-51x+620=0$

When solved you should get $x=31$ or $x=20$. I'll let you test these values. Only one is correct.

The book answer is incorrect. 21 is not a possible solution for x.

--Chris

3. Its amazing how i seem to miss the tiniest details....

Thanks for the help