Question

Find the sum of the squares of the following: $\frac{\sqrt{3}}{\sqrt{2}+1},\frac{\sqrt{3}}{\sqrt{2}-1},\frac{\sqrt{2}}{\sqrt{3}}$

• A. $\frac{56}{4}$
• B. $18\frac{2}{3}$
• C. $2\frac{18}{3}$
• D. $3\frac{18}{2}$

Answer 1

The correct option is B $18\frac{2}{3}$

$i).(\frac{\sqrt{3}}{\sqrt{2}+1}{)}^2=\frac{(\sqrt{3}{)}^2}{(\sqrt{2}{)}^2+(1{)}^2+2\sqrt{2}}=\frac{3}{2+1+2\sqrt{2}}=\frac{3}{3+2\sqrt{2}}\left(ii\right).(\frac{\sqrt{3}}{\sqrt{2}-1}{)}^2=\frac{(\sqrt{3}{)}^2}{(\sqrt{2}{)}^2+(1{)}^2-2\sqrt{2}}=\frac{3}{2+1-2\sqrt{2}}=\frac{3}{3-2\sqrt{2}}iii)(\frac{\sqrt{2}}{\sqrt{3}}{)}^2=\frac{2}{3}\therefore (\frac{\sqrt{3}}{\sqrt{2}+1}{)}^2+(\frac{\sqrt{3}}{\sqrt{2}-1}{)}^2+(\frac{\sqrt{2}}{\sqrt{3}}{)}^2=[\frac{3}{3+2\sqrt{2}}+\frac{3}{3-2\sqrt{2}}]+\frac{2}{3}=\frac{3(3-2\sqrt{2})+3(3+2\sqrt{2})}{(3{)}^2-(2\sqrt{2}{)}^2}+\frac{2}{3}=\frac{9-6\sqrt{2}+9+6\sqrt{2}}{9-8}+\frac{2}{3}=18+\frac{2}{3}=\frac{56}{3}=18\frac{2}{3}$