Question

Find the square root of $3+\sqrt{2}$.

Answer 1

Let $\sqrt{3+\sqrt{2}}=\sqrt{p}+\sqrt{q}$
$3+\sqrt{2}=p+q+2\sqrt{pq}$ [By squaring both sides]
$\therefore p+q=3$ .....(i) [By equating the parts]
$\Rightarrow 2\sqrt{pq}=\sqrt{2}$ ....(ii)
$\Rightarrow 4pq=2$ ..........(iii) [By squaring both sides]

$(p-q{)}^2=(p+q{)}^2-4pq$
$\Rightarrow$ $(p-q{)}^2=9-2$

$\Rightarrow (p-q{)}^2=7$
$\Rightarrow$ $p-q=\sqrt{7}$ ...........(iv)
$\because p+q=3$ [By $e{q}^n\left(i\right)$]
$\because p=\frac{1}{2}(3+\sqrt{7})$ [On adding (i) & (iv)]
$\Rightarrow q=\frac{1}{2}(3-\sqrt{7})$ [On subtracting (i) & (iv)]
$\therefore$ $\sqrt{3+\sqrt{2}}=\pm \frac{1}{\sqrt{2}}(\sqrt{3+\sqrt{7}}+\sqrt{3-\sqrt{7}})$