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The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Given:
The difference of the squares of two numbers 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 $x^2=8(18)=144$
$x^2=(12)^2$
$x=\pm 12$
The required numbers are $12, 18$ or $-12, 18$.
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