Question

Prove:$\sqrt{23+\sqrt{528}}=2\sqrt{3}+\sqrt{11}$

Answer 1

Prove : $\sqrt{23+\sqrt{528}}=2\sqrt{3}+\sqrt{11}$

$LHS:-\sqrt{23+\sqrt{528}}$

$=\sqrt{23+\sqrt{4^4\times 3\times 11}}$

$=\sqrt{23+2^2\sqrt{3\times 11}}$

$=\sqrt{23+4\sqrt{33}}$

$=\sqrt{12+11+4\sqrt{33}}$

$=\sqrt{12+2\sqrt{33}+11}$

$=\sqrt{(\sqrt{42}{)}^2+2\times \sqrt{12}\times \sqrt{11}+{\sqrt{11}}^2}$

$=\sqrt{(\sqrt{12}+\sqrt{11}{)}^2}$ $(\because a^2+2ab+b^2-(a+b{)}^2)$

$=\sqrt{12}+\sqrt{11}$

$=\sqrt{4\times 3}+\sqrt{11}$

$=2\sqrt{3}+\sqrt{11}$

$LHS=RHS$