Find the consecutive positive integers, sum of whose square is 365.
There is difference of 1 in consecutive positive integers.
Let first integer =x
Second integer =x+1
Also given that
Sum of squares =365
x2+(x+1)2=365
⇒x2+x2+1+2x=365
⇒2x2+1+2x=365
⇒2x2+1+2x−365=0
⇒2x2+2x−364=0
⇒2(x2+x−182)=0
⇒x2+x−182=0
⇒x2+14x−13x−182=0
⇒x(x+14)−13(x+14)=0
⇒(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.