Question

The sum of the squares of two natural numbers is $233$ and one of the numbers is $3$ less than twice the other number. The smaller number is:

Answer 1

$\Rightarrow$ Let the number be $x$

$\Rightarrow$ Then the other number $=2x-3$

According to the question,

$\Rightarrow$ $x^2+(2x-3{)}^2=233$

$\Rightarrow$ $x^2+4x62+9-12x=233$

$\Rightarrow$ $5x^2-12x-224=0$ ------ ( 1 )

$\Rightarrow$ Here, $a=5,b=-12,c=-224$

$\Rightarrow$ $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{-(-12)\pm \sqrt{(-12{)}^2-4(5\left)\right(-224)}}{2\left(5\right)}$

$=\frac{12\pm \sqrt{144+4480}}{10}$

$=\frac{12\pm \sqrt{4624}}{10}$

$=\frac{12\pm 68}{10}$

$\Rightarrow$ $\frac{12+68}{10}$ and $x=\frac{12-68}{10}$

$\Rightarrow$ $x=8$ and $x=\frac{-28}{5}$

$x$ can not be negative.

$\therefore$ $x=8$

$\Rightarrow$ Other number $=2x-3=2\times 8-3=15$

$\therefore$ Here, smaller number is $8$.