An algebra problem by Martin Nikolov

$\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}$

$=\frac{1}{\sqrt{2}}(\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2)$

$=\frac{1}{\sqrt{2}} [(\sqrt{7}+1)-(\sqrt{7}-1)-2]=\boxed{0}$

First let $S=\sqrt{4+\sqrt7}-\sqrt{4-\sqrt7}$ , so $S^2=4+\sqrt7+4-\sqrt7-2\sqrt{4^2-7}=2$ . Therefore, $S=\sqrt2$ , so the total sum is $\sqrt2-\sqrt2=0$

$\sqrt{4 + \sqrt{7}} = \sqrt{\frac{7}{2} + \frac{1}{2} + 2.\frac{\sqrt{7}.1}{\sqrt{2}.\sqrt{2}}}$

given expression

$= \sqrt{ (\sqrt{\frac{7}{2}} + \sqrt{\frac{1}{2}})^{2}} - \sqrt{ (\sqrt{\frac{7}{2}} - \sqrt{\frac{1}{2}})^{2}} - \sqrt{2}$

$= \frac{2}{\sqrt{2}} - \sqrt{2}$

x= √(4+√(7)) - √(4-√(7)) - √(2) ⇒
( x+√(2))² = ( √(4+√(7)) - √4-√7)) )² ⇒
x²+2√(2)x+2=4+√ (7)-2√(16-7) +4-√(7) ⇒
x²+2√(2)x+2 -2=4-2√(9)+4 -2 x²+2√(2)x = 6 - 2(±3)
1.) x² +2√(2)x = 6-(6)
2.) x² + 2√(2)x = 6+(6) 1.) x = -2√(2) or 0 * * 2.) x = -(2)± √(14)

I got four solutions but I pick the only integer I got. sorry for the weird confusions.

\sqrt(4+\sqrt(7))-\sqrt(4-\sqrt(7))-\sqrt(2) =(\frac{1}{\sqrt(2)})\times \sqrt(7 +2times (\sqrt(7))+1)-\sqrt(7+1-2times (\sqrt(7)))-2

ans = 0 sqrt(4+sqrt(7)) = (sqrt(7) + 1)/sqrt(2) sqrt( 4 - sqrt(7) ) = (sqrt(7) - 1)/sqrt(2) sqrt( 4 + sqrt(7) ) - sqrt( 4 - sqrt(7) ) = sqrt(2) sqrt(2) - sqrt(2) = 0