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The sum of the squares of two numbers is 233 and one of the numbers is 3 less than twice the other number. Find the numbers.
Given:
The sum of the squares of two numbers is 233 and one of the numbers is 3 less than twice the other number.
To do:
We have to find the numbers.
Solution:
Let one of the numbers be $x$.
This implies,
The other number $=2x-3$.
According to the question,
$x^2+(2x-3)^2=233$
$x^2+4x^2-12x+9=233$
$5x^2-12x+9-233=0$
$5x^2-12x-224=0$
Solving for $x$ by factorization method, we get,
$5x^2-12x-224=0$
$5x^2-40x+28x-224=0$
$5x(x-8)+28(x-8)=0$
$(5x+28)(x-8)=0$
$5x+28=0$ or $x-8=0$
$x=8$ (Since $5x+28≠0$)
$2x-3=2(8)-3=16-3=13$
Therefore, the required numbers are $8$ and $13$.
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