# Math Help - radicals HELP!!!!! URGENT!!!!!

sq. rt. (x-7) + sq. rt. (x) =4

2. Originally Posted by iheartthemusic29
sq. rt. (x-7) + sq. rt. (x) =4
$\sqrt{x - 7} + \sqrt{x} = 4$

The trick here is to divide and conquer. You want to isolate one of the square roots on one side of the equation. So:
$\sqrt{x - 7} = 4 - \sqrt{x}$ <-- Square both sides

$x - 7 = \left ( 4 - \sqrt{x} \right ) ^2$

$x - 7 = 16 - 8\sqrt{x} + x$

Now do it again:
$\sqrt{x} =\frac{23}{8}$

$x = \left ( \frac{23}{8} \right ) ^2 = \frac{529}{64}$

Always always always check your solutions in the original equation:
$\sqrt{\frac{529}{64} - 7} + \sqrt{\frac{529}{64}} = 4$

$\sqrt{\frac{81}{64}} + \frac{23}{8} = 4$

$\frac{9}{8} + \frac{23}{8} = 4$

$\frac{32}{8} = 4$

$4 = 4$ (Check!)

-Dan