Question:
Find two consecutive numbers whose squares have the sum 85.
Solution:
Let two consecutive numbers be $x$ and $(x+1)$
Then according to question
$x^{2}+(x+1)^{2}=85$
$x^{2}+x^{2}+2 x+1=85$
$2 x^{2}+2 x-85+1=0$
$2 x^{2}+2 x-84=0$
$x^{2}+x-42=0$
$x^{2}+7 x-6 x-42=0$
$x(x+7)-6(x+7)=0$
$(x+7)(x-6)=0$
$(x+7)=0$
$x=-7$
Or
$(x-6)=0$
$x=6$
Since, x being a number,
Therefore,
When $x=-7$ then
$x+1=-7+1$
$=-6$
And when $x=6$ then
$x+1=6+1$
$=7$
Thus, two consecutive number be either 6,7 or $-6,-7$
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