Find two consecutive numbers whose squares have the sum 85.

Given:

Two consecutive numbers whose squares have the sum 85.

 To do:

We have to find the two numbers.

Solution:

Let the two consecutive numbers be $x$ and $x+1$.

This implies,

$x^2+(x+1)^2=85$

$x^2+x^2+2x+1=85$    (Since $(a+b)^2=a^2+2ab+b^2$)

$2x^2+2x+1-85=0$

$2x^2+2x-84=0$

$2(x^2+x-42)=0$

$x^2+x-42=0$

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

$x^2+7x-6x-42=0$

$x(x+7)-6(x+7)=0$

$(x+7)(x-6)=0$

$x+7=0$ or $x-6=0$

$x=-7$ or $x=6$

If $x=-7$, $x+1=-7+1=-6$

The two consecutive integers are $-7$ and $-6$.

If $x=6$, $x+1=6+1=7$

The two consecutive integers are $6$ and $7$.

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Updated on: 10-Oct-2022

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