Question

If $\sqrt{18-2\sqrt{45}}=\sqrt{x}-\sqrt{y}$, then find values of $x$ and $y$.

Answer 1

Given, $\sqrt{18-2\sqrt{45}}=\sqrt{x}-\sqrt{y}$

Squaring both sides, we get

$18-2\sqrt{45}=x+y-2\sqrt{xy}$

On comparing, we get,

$x+y=18$ .....(i)

and $\sqrt{45}=\sqrt{xy}$ ......(ii)

From eqn (ii), we get

$45=xy$

We know $(x-y{)}^2=(x+y{)}^2-4xy$

$=(18{)}^2-4\times 45$

$=324-180=144$

Thus $\sqrt{(x-y{)}^2}=\sqrt{144}$

$\Rightarrow x-y=\pm 12$ .....(iii)

Solving eqn (i) and eqn (iii) simultaneously, we get

$x=15,y=3$ or $x=3,y=15$