Question

The value of y varies inversely as the square of x, and y=4, when x=3.

Answer 1

The value of x when y = 9 is x = 2 or x = -2

Solution:

Given that value of y varies inversely as the square of x, and y=4, when x=3.

Therefore the initial statement is:

$y \propto \frac{1}{x^{2}}$

To convert to an equation, multiply by k, the constant  of variation

$y = k \times \frac{1}{x^2}$

$y = \frac{k}{x^2}$  --- eqn 1

Given that,

y = 4 when x = 3

Now find value of k

$4 = \frac{k}{3^2}\\\\4 \times 9 = k\\\\k = 36$

Find the value of x when y = 9

x = ?

y = 9

From eqn 1,

$9 = \frac{k}{x^2}\\\\9 = \frac{36}{x^2}\\\\x^2 = 4\\\\x = \pm 2$

Thus value of x is found