Question

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

Answer 1

$\Rightarrow$ Let the smaller positive number be $x$ and larger number be $y$

$\therefore$ $x^2=4y$ ------ ( 1 ) [ Given ]

$\Rightarrow$ $y^2-x^2=45$ [ Given ]

$\Rightarrow$ $y^2-4y=45$ [ From ( 1 ) ]

$\Rightarrow$ $y^2-4y-45=0$

$\Rightarrow$ $y^2-9y+5y-45=0$

$\Rightarrow$ $y(y-9)+5(y-9)=0$

$\Rightarrow$ $(y-9)(y+5)=0$

$\therefore$ $y=9$ and $y=-5$

$\Rightarrow$ $y=-5$ is not possible because $y$ is positive number.

$\therefore$ $y=9$

Substituting value of $y$ in equation ( 1 ),

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

$x^2=36$

$\therefore$ $x=6$

$\Rightarrow$ The required two numbers are $9$ and $6$

$\Rightarrow$ Sum of the numbers $=9+6=15$