Question

The sum of the squares of two natural numbers is $34$. If the first number is one less than twice the second number, find the numbers.

Answer 1

Step 1: Consider the given details

Let the first and second natural numbers be $x$ and $y$ respectively.

Given that the sum of the squares of two natural numbers is $34$.

$x^2+y^2=34...\left(i\right)$

Also given that, the first number is one less than twice the second number.

$x=2y-1...\left(ii\right)$

Substitute $\left(ii\right)$ in $\left(i\right)$ and simplify as

$$\begin{gathered} {(2y-1)}^2+y^2=34 \\ \Rightarrow 4y^2-4y+1+y^2=34 \\ \Rightarrow 5y^2-4y+1=34 \\ \Rightarrow 5y^2-4y-33=0 \end{gathered}$$

Step 2: Solve the quadratic equation by substituting the values in the quadratic formula

The quadratic equation formed is $5y^2-4y-33=0$.

Compare the above equation with the standard form of the quadratic equation which is $a{x}^2+bx+c=0$.

So,

$$\begin{gathered} a=5 \\ b=-4 \\ c=-33 \end{gathered}$$

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

After substituting,

$\Rightarrow y=\frac{-(-4)\pm \sqrt{{(-4)}^2-4\times 5\times -33}}{2\times 5}$

On simplifying,

$$\begin{gathered} \Rightarrow y=\frac{4\pm \sqrt{16+660}}{10} \\ \Rightarrow y=\frac{4\pm \sqrt{676}}{10} \\ \Rightarrow y=\frac{4\pm 26}{10} \end{gathered}$$

The above equation can be written as,

$\Rightarrow y=\frac{4+26}{10}$ and $y=\frac{4-26}{10}$

$\Rightarrow y=\frac{30}{10}$ and $y=\frac{-22}{10}$

$\therefore y=3$ and $y=\frac{-22}{10}$

Here only $y=3$ is valid since it's given that the numbers are natural numbers.

From $\left(ii\right)$,

$x=2y-1=6-1=5$

Hence, the required numbers are $3$ and $5$.