# Suppose that y varies inversely with the square root of x and y=50 when x=4, how do you find y when x=5?

May 28, 2015

If $y$ varies inversely with $\sqrt{x}$
then
$y \cdot \sqrt{x} = c$ for some constant $c$

Given $\left(x , y\right) = \left(4 , 50\right)$ is a solution to this inverse variation
then
$50 \cdot \sqrt{4} = c$
$\rightarrow c = 100 \textcolor{w h i t e}{\text{xxxxxxxxxx}}$ (see note below)
and the inverse variation equation is
$y \cdot \sqrt{x} = 100$

When $x = 5$ this becomes
$y \cdot \sqrt{5} = 100$

$\sqrt{5} = \frac{100}{y}$

$5 = {10}^{4} / {y}^{2}$

$y = \sqrt{5000} = 50 \sqrt{2}$

Note: I have interpreted "y varies inversely with the square root of x" to mean the positive square root of x (i.e. $\sqrt{x}$) which also implies that y is positive. If this is not the intended case, the negative version of y would also need to be allowed.