I am not sure how to approach this question, can I get some help? This is an inverse proportion question.
$y$ is inversely proportional to the square of $x$.
Find the percentage change of $x$ when $y$ is decreased by $36\%$ .
Not sure where to continue after my working:
Old $y$ value = $1 y$
New $y$ value = $0.64 y$
$$y = \frac{k}{x^2}\tag{1}$$
$$0.64y = \frac{k}{x^2}\tag{2}$$
Let the initial value of $x$ be $x_1$ and the value after the change in $y$ be $x_2$. Continuing from where you left off, $$y=\frac{k}{{x_1}^2}\tag1$$ $$0.64y=\frac{k}{{x_2}^2}\tag2$$ Dividing $(2)$ by $(1)$, $$0.64=\frac{{x_1}^2}{{x_2}^2}$$ $$0.8=\frac{x_1}{x_2}$$ $$x_1=0.8x_2$$ or $$x_2=1.25x_1$$
Thus, the value of $x$ increases by $25\%$