Find two natural numbers which differ by 3 and whose squares have the sum 117.

Given:

Two natural numbers differ by 3 and their squares have the sum 117.

 To do:

We have to find the numbers.

Solution:

Let the two natural numbers be $x$ and $x+3$.

According to the question,

$(x)^2+(x+3)^2=117$

$x^2+x^2+6x+9=117$

$2x^2+6x+9-117=0$

$2x^2+6x-108=0$

$2(x^2+3x-54)=0$

$x^2+3x-54=0$

Solving for $x$ by factorization method, we get,

$x^2+9x-6x-54=0$

$x(x+9)-6(x+9)=0$

$(x+9)(x-6)=0$

$x+9=0$ or $x-6=0$

$x=-9$ or $x=6$

$-9$ is not a natural number. Therefore, $x=6$.

$x+3=6+3=9$

The required natural numbers are $6$ and $9$.

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Updated on: 10-Oct-2022

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