Question

The sum of the squares of two natural numbers is $52$. If the first number is $8$ less than twice the second number, find the numbers.

Answer 1

Step 1: Forming a quadratic equation from the given details

Let the two natural numbers be $x$ and $y$.

Given that the first number is $8$ less than twice of the second number.

So, $x=2y-8$.

Also given that the sum of the squares is $52$.

So, $y^2+{(2y-8)}^2=52$.

Simplifying it further,

$$\begin{gathered} \Rightarrow y^2+(4y^2-32y+64)=52 \\ \Rightarrow y^2+4y^2-32y+64-52=0 \\ \Rightarrow 5y^2-32y+12=0 \end{gathered}$$

Step 2: Using the quadratic formula to find the numbers

The quadratic formula is $y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

The standard form of the quadratic equation which is $a^2+bx+c=0$.

Comparing with $5y^2-32y+12=0$,

$$\begin{gathered} a=5 \\ b=-32 \\ c=12 \end{gathered}$$

After substitution,

$\Rightarrow y=\frac{-(-32)\pm \sqrt{{(-32)}^2-4\times 5\times 12}}{2\times 5}$

On simplifying,

$$\begin{gathered} \Rightarrow y=\frac{32\pm \sqrt{1024-240}}{10} \\ \Rightarrow y=\frac{32\pm \sqrt{784}}{10} \\ \Rightarrow y=\frac{32\pm 28}{10} \end{gathered}$$

This can be written as,

$\Rightarrow y=\frac{32+28}{10}$ and $y=\frac{32-28}{10}$

$\Rightarrow y=\frac{60}{10}$ and $y=\frac{4}{10}$

Therefore $y=6$ and $y=\frac{4}{10}$.

Here $y=6$ will be considered since the given number is a natural number.

Hence the first number will be $4$ and the second number will be $6$.