Question

If $\sqrt{13-x\sqrt{10}}=\sqrt{8}+\sqrt{5}$ , then $x=$ .

Answer 1

Given: $\sqrt{13-x\sqrt{10}}=\sqrt{8}+\sqrt{5}$

Squaring on both sides,

$(\sqrt{13-x\sqrt{10}}{)}^2=(\sqrt{8}+\sqrt{5}{)}^2$

$13-x\sqrt{10}=(\sqrt{8}{)}^2+(\sqrt{5}{)}^2+2\sqrt{5}\times \sqrt{8}$

$13-x\sqrt{10}=8+5+2\sqrt{40}$

$13-x\sqrt{10}=13+2\sqrt{40}$

$13-x\sqrt{10}=13+2\times 2\sqrt{10}$

$13-x\sqrt{10}=13+4\sqrt{10}$

Comparing above equation we get,

$x=-4$