Question

If the square root of $\sqrt{50}+\sqrt{48}$ = $2^{1/m}(\sqrt{3}+\sqrt{2})$. find m

Answer 1

Solving $L.H.S.$

$\sqrt{\sqrt{50}+\sqrt{48}}$

$\sqrt{\frac{\sqrt{2}}{\sqrt{2}}(\sqrt{50}+\sqrt{48)}}$

$\frac{\sqrt{(\sqrt{100}+\sqrt{96)}}}{2^{\frac{1}{4}}}$

$\frac{\sqrt{(10+2\sqrt{24)}}}{2^{\frac{1}{4}}}$

$\frac{\sqrt{(6+4+2\sqrt{24)}}}{2^{\frac{1}{4}}}$

$\frac{\sqrt{(\sqrt{(6{)}^2}+\sqrt{(4{)}^2}+2\times \sqrt{6}\times \sqrt{4}}}{2^{\frac{1}{4}}}$ Using $a^2+b^2+2ab=(a+b{)}^2$

$\frac{(\sqrt{6}+\sqrt{4})}{2^{\frac{1}{4}}}$

$\frac{\sqrt{2}(\sqrt{3}+\sqrt{2})}{2^{\frac{1}{4}}}$

$2^{\frac{1}{4}}(\sqrt{3}+\sqrt{2})$

According To Question $L.H.S.=R.H.S.$

$2^{\frac{1}{4}}(\sqrt{3}+\sqrt{2})$$=2^{1/m}(\sqrt{3}+\sqrt{2})$

$2^{1/4}=2^{1/m}$

$ComparingPowers$

$\frac{1}{4}=\frac{1}{m}$

$m=4$

Answer is $4$