Question

The difference between 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

Step 1. Convert statements into equations :

Assume that the numbers are $x\mathrm{and}y$.

The difference between the squares of two natural numbers is $45$.

So, the equation formed is:

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

Here, $y$ is the smaller number and the square of $y$ is $4$ times $x$.

So, the equation formed is:

$y^2=4x$

Step 2. Solve the equation obtained :

Substitute $y^2=4x$ into $x^2-y^2=45$ and solve for $x$:

$\begin{array}{rcl}{x}^2-4y& =& 45\\ \Rightarrow x^2-4y-45& =& 0\\ & \Rightarrow & x^2-9x+5x-45=0\\ \Rightarrow \left(x-9\right)\left(x+5\right)& =& 0\end{array}$

A number is a natural number.

So, $x\ne 5\mathrm{and}x=9$.

Substitute $x=9$ into $y^2=4x$ and solve for $y$:

$\begin{array}{rcl}{y}^2& =& 4\times 9\\ & \Rightarrow & y^2=36\end{array}$

The value of $y$ cannot be negative.

So, $y=6$.

Hence, the numbers are $x=9,y=6$.