Question

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

Answer 1

Let the smaller natural number be x and larger natural number be y

Hence $x^2=4y$ → (1)

Given $y^2–x^2=45$

⇒ $y^2–4y=45$

⇒$y^2–4y–45=0$

⇒ $y^2–9y+5y–45=0$

⇒ $y(y–9)+5(y–9)=0$

⇒$(y–9)(y+5)=0$

⇒$(y–9)=0or(y+5)=0$

∴$y=9ory=-5$

But y is natural number, hence y ≠ - 5

Therefore, y = 9

Equation (1) becomes,

$x^2=4\left(9\right)=36$

∴ x = 6

Thus the two natural numbers are 6 and 9.