Question

Find the consecutive positive integers, sum of whose square is $365$.

Answer 1

There is difference of $1$ in consecutive positive integers.

Let first integer $=x$

Second integer $=x+1$

Also given that

Sum of squares $=365$

$x^2+(x+1{)}^2=365$

$\Rightarrow x^2+x^2+1+2x=365$

$\Rightarrow 2x^2+1+2x=365$

$\Rightarrow 2x^2+1+2x-365=0$

$\Rightarrow 2x^2+2x-364=0$

$\Rightarrow 2(x^2+x-182)=0$

$\Rightarrow x^2+x-182=0$

$\Rightarrow x^2+14x-13x-182=0$

$\Rightarrow x(x+14)-13(x+14)=0$

$\Rightarrow (x-13)(x+14)=0$

$x-13=0$ $x+14=0$

$x=13$ $x=-14$

So, the root of the equation are $x=13$ & $x=-14$

Since we have to find consecutive positive number

We take $x=13$

First number $=x=13$

Second number $=x+1=13+1=14$.