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:
[tex]y \propto \frac{1}{x^{2}}[/tex]
To convert to an equation, multiply by k, the constant of variation
[tex]y = k \times \frac{1}{x^2}[/tex]
[tex]y = \frac{k}{x^2}[/tex] --- eqn 1
Given that,
y = 4 when x = 3
Now find value of k
[tex]4 = \frac{k}{3^2}\\\\4 \times 9 = k\\\\k = 36[/tex]
Find the value of x when y = 9
x = ?
y = 9
From eqn 1,
[tex]9 = \frac{k}{x^2}\\\\9 = \frac{36}{x^2}\\\\x^2 = 4\\\\x = \pm 2[/tex]
Thus value of x is found
