Question

Show that: $\sqrt{19-4\sqrt{12}}+\sqrt{7+4\sqrt{3}}=3$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\sqrt{19-4\sqrt{12}}+\sqrt{7+4\sqrt{3}} \\ =\sqrt{19-8\sqrt{3}}+\sqrt{7+4\sqrt{3}} \\ =\sqrt{\left(16+3\right)-8\sqrt{3}}+\sqrt{\left(4+3\right)+4\sqrt{3}} \\ =\sqrt{{\left(4\right)}^2+{\left(\sqrt{3}\right)}^2-2\times 4\times \sqrt{3}}+\sqrt{{\left(2\right)}^2+{\left(\sqrt{3}\right)}^2+2\times 2\times \sqrt{3}} \\ =\sqrt{{\left(4-\sqrt{3}\right)}^2}+\sqrt{{\left(2+\sqrt{3}\right)}^2} \\ =4-\sqrt{3}+2+\sqrt{3} \\ =6 \\ \mathrm{RHS}=6 \\ \mathrm{LHS}=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$