The difference of the squares of two positive integers is 180. The square of the smaller number is 8 times the larger, find the numbers.

Given:

The difference of the squares of two positive integers is 180. The square of the smaller number is 8 times the larger.

 To do:

We have to find the numbers.

Solution:

Let the two numbers be $x$ and $y$ in which $x$ is the smaller number.

According to the question,

$y^2-x^2=180$ and $x^2=8y$

$y^2-x^2=180$

$y^2-8y=180$

$y^2-8y-180=0$

Solving for $y$ by factorization method,

$y^2-18y+10y-180=0$

$y(y-18)+10(y-18)=0$

$(y-18)(y+10)=0$

$y-18=0$ or $y+10=0$

$y=18$ or $y=-10$

$-10$ is not a positive integer. Therefore, $y=18$.

$y=18$, then $x^2=8(18)=144$

$x^2=(12)^2$

$x=12$

The required numbers are $18$ and $12$ respectively.

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Updated on: 10-Oct-2022

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