Question

Show that : $\sqrt{19-6\sqrt{10}}=\sqrt{10}-3$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\sqrt{19-6\sqrt{10}} \\ =\sqrt{10+9-6\sqrt{10}} \\ =\sqrt{{\left(\sqrt{10}\right)}^2+{\left(3\right)}^2-2\times \sqrt{10}\times 3} \\ =\sqrt{{\left(\sqrt{10}-3\right)}^2} \\ =\sqrt{10}-3 \\ \mathrm{RHS}=\sqrt{10}-3 \\ \mathrm{LHS}=\mathrm{RHS} \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$