Question

If $\sqrt{13-a\sqrt{10}}=\sqrt{8}+\sqrt{5}$ then $a=$

Answer 1

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

Squaring on both sides,

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

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

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

$13-a\sqrt{10}=13+2\times \sqrt{4\times 10}$

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

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

Comparing we get,

$a=-4$

$\therefore \left(c\right)$ is the Correct Option