Question

$y$ varies directly with square root of $x$ and $y=9$ when $x=9$. Find the value of $x$ when $y=6$

Answer 1

As per question, $y\propto \sqrt{x}\Rightarrow y=k\sqrt{x}\cdots \cdots \cdots \cdots$ (1)
(where $k=$ non-zero variation constant)
When $x=9,y=9,\therefore 9=k\sqrt{9}\text{[by(1)]}$
or, $9=k.3$ or, $k=\frac{9}{3}=3$.
$\therefore$ from (1) we get, $y=3\sqrt{x}....\left(2\right)$
Now, when $y=6$, from (2) we get, $6=3\sqrt{x}$
or, $\sqrt{x}=\frac{6}{3}=2$
or, $x=(2{)}^2=4$ [Squaring both the sides]
$\therefore$ The required value of $x=4$.