Question

The difference of the squares of two positive number is $45$. The square of the smaller number is four times the larger number. Find the numbers.

Answer 1

Let $x$ and $y$ be the two numbers such that $x>y$

Given$x^2-y^2=45$

And $y^2=4x$

$\Rightarrow x^2-4x-45=0$

$\Rightarrow x^2-9x+5x-45=0$

$\Rightarrow x(x-9)+5(x-9)=0$

$\Rightarrow (x+5)(x-9)=0$

$\therefore x+5=0$, $x-9=0$

$\therefore x=-5,9$

For $x=-5,y^2=4\times -5=-20$ which is not a valid solution

For $x=9.y^2=4\times 9=36$ or $y=\pm 6$

From the above condition $x>y$

$\therefore 9>6$ is the only solution

Hence, $9,6$ are the required numbers