Question

Find the square root of following surd :
$6+4\sqrt{2}$

Answer 1

Let $\sqrt{6+4\sqrt{2}}=\pm (\sqrt{a}+\sqrt{b})$

Squaring both sides,

$6+4\sqrt{2}=a+b+2\sqrt{ab}$

Comparing both sides,

$a+b=6$ ........ $\left(1\right)$

$4\sqrt{2}=2\sqrt{ab}$

or $8=ab\Rightarrow a=\frac{8}{b}$ ............ $\left(2\right)$

$\therefore$ From $\left(1\right)$ and $\left(2\right)$

$\frac{8}{b}+b=6$

or, $8+b^2=6b$

or, $b^2-6b+8=0$

or, $b^2-4b-2b+8=0$

or, $b(b-4)-2(b-4)=0\Rightarrow (b-4)(b-2)=0$

$\therefore b=2,2$

$\therefore a=2,4$

$\therefore$ Squaring root is $\pm (\sqrt{2}+\sqrt{4})=\pm (2+\sqrt{2})$ Ans