Question

Solve the following equation:
$\sqrt{x-\sqrt{x-2}}+\sqrt{x+\sqrt{x-2}}=3$

Answer 1

$\sqrt{x-\sqrt{x-2}}+\sqrt{x+\sqrt{x-2}}=3$

squaring both sides

$\Rightarrow x-\sqrt{x-2}+x+\sqrt{x-2}+2\sqrt{x-\sqrt{x-2}}\sqrt{x+\sqrt{x-2}}=9$

$\Rightarrow 2x+2\sqrt{x-\sqrt{x-2}}+\sqrt{x+\sqrt{x-2}}=9$

$2\sqrt{x-\sqrt{x-2}}+\sqrt{x+\sqrt{x-2}}=9-2x$ squaring the 2 sides

$4\left[(x-\sqrt{x-2})(x+\sqrt{x-2})\right]=81+4x^2-36x$

$4(x^2-(x-2))=4x^2-36x+81$

$4(x^2-x+2)=4x^2-36x+81$

$4x^2-4x+8=4x^2-36x+81$

$-4x+36x=81-8$

$32x=73$

$\Rightarrow x=\frac{73}{32}.$

Hence, the answer is $\frac{73}{32}.$