Question

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

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \sqrt{13-x\sqrt{10}}=\sqrt{8}+\sqrt{5} \\ \mathrm{Now}, \\ \sqrt{13-x\sqrt{10}}=\sqrt{8}+\sqrt{5} \\ \Rightarrow \sqrt{13-x\sqrt{10}}=\sqrt{{\left(\sqrt{8}+\sqrt{5}\right)}^2} \\ \Rightarrow \sqrt{13-x\sqrt{10}}=\sqrt{{\left(\sqrt{8}\right)}^2+{\left(\sqrt{5}\right)}^2+2\left(\sqrt{8}\right)\left(\sqrt{5}\right)}\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:\hspace{0.17em}{\left(a+b\right)}^2=a^2+b^2+2ab\right) \\ \Rightarrow \sqrt{13-x\sqrt{10}}=\sqrt{8+5+2\left(2\sqrt{2}\right)\left(\sqrt{5}\right)} \\ \Rightarrow \sqrt{13-x\sqrt{10}}=\sqrt{13+4\sqrt{10}} \\ \Rightarrow -x=4 \\ \Rightarrow x=-4 \end{gathered}$$


Hence, if $\sqrt{13-x\sqrt{10}}=\sqrt{8}+\sqrt{5},\mathrm{then}x=\underline{-4}.$