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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Updated on: 10-Oct-2022

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