The difference of the squares of two numbers is 45. The square of the smaller number is 4 times the larger number. Determine the numbers.

Let the larger number be x .

Given, the square of the smaller number is 4 times the larger number.

Hence, square of smaller number = 4x

Given, the difference of squares of two numbers is = 45

$\Rightarrow x^2 - 4x = 45 \\[0.5em] \Rightarrow x^2 - 4x - 45 = 0 \\[0.5em] \Rightarrow x^2 - 9x + 5x - 45 = 0 \\[0.5em] \Rightarrow x(x - 9) + 5(x - 9) = 0 \\[0.5em] \Rightarrow (x + 5)(x - 9) = 0 \\[0.5em] \Rightarrow x + 5 = 0 \text{ or } x - 9 = 0 \\[0.5em] \Rightarrow x = -5 \text{ or } x = 9$

If x = -5 , $\sqrt{4x} = \sqrt{-20}$ , which is not valid as there is no real value of square root of a negative value.

∴ x = 9 , $\sqrt{4x} = \sqrt{36} = 6 \text{ or } -6$.

Hence, the two numbers are 9, 6 or 9, -6.