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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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