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Find two consecutive positive integers, sum of whose squares is 365.
Given:
Two consecutive numbers whose squares have the sum 365.
To do:
We have to find the two numbers.
Solution:
Let the two consecutive numbers be $x$ and $x+1$.
This implies,
$x^2+(x+1)^2=365$
$x^2+x^2+2x+1=365$ (Since $(a+b)^2=a^2+2ab+b^2$)
$2x^2+2x+1-365=0$
$2x^2+2x-364=0$
$2(x^2+x-182)=0$
$x^2+x-182=0$
Solving for $x$ by factorization method, we get,
$x^2+14x-13x-182=0$
$x(x+14)-13(x+14)=0$
$(x+14)(x-13)=0$
$x+14=0$ or $x-13=0$
$x=-14$ or $x=13$
If $x=13$, $x+1=13+1=14$
The two consecutive positive integers are $13$ and $14$.
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