Question

Question 9. If $\sqrt{13-x\sqrt{10}}=\sqrt{8}+\sqrt{5}$ then find the value of x.
(A) - 5
(B) - 6
(C) - 4
(D) - 2

Answer 1

$$\begin{gathered} \sqrt{13-\mathrm{x}\sqrt{10}}=\sqrt{8}+\sqrt{5} \\ \Rightarrow \sqrt{8+5-\mathrm{x}\sqrt{10}}=2\sqrt{2}+\sqrt{5} \\ \Rightarrow \sqrt{{\left(2\sqrt{2}\right)}^2+{\left(\sqrt{5}\right)}^2-\mathrm{x}\times 2\sqrt{2}\times \sqrt{5}}=2\sqrt{2}+\sqrt{5} \\ \mathrm{Taking}\mathrm{x}=-2,\mathrm{we}\mathrm{get} \\ \sqrt{{\left(2\sqrt{2}\right)}^2+{\left(\sqrt{5}\right)}^2-\left(-2\right)\times 2\sqrt{2}\times \sqrt{5}}=2\sqrt{2}+\sqrt{5} \\ \Rightarrow \sqrt{{\left(2\sqrt{2}\right)}^2+{\left(\sqrt{5}\right)}^2+2\times 2\sqrt{2}\times \sqrt{5}}=2\sqrt{2}+\sqrt{5}\left[\mathbf{as}\mathbf{,}\mathbf{}{\mathbf{a}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{b}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{ab}\mathbf{}\mathbf{=}\mathbf{}{\left(\mathbf{a}\mathbf{+}\mathbf{b}\right)}^{\mathbf{2}}\right] \\ \Rightarrow \sqrt{{\left(2\sqrt{2}+\sqrt{5}\right)}^2}=2\sqrt{2}+\sqrt{5} \\ \Rightarrow 2\sqrt{2}+\sqrt{5}=2\sqrt{2}+\sqrt{5} \\ \mathrm{Hence},\mathrm{x}=-2 \end{gathered}$$