Question

The sum of the squares of the three positive consecutive numbers is 365.find the numbers

Answer 1

Let the consecutive numbers be x , x+1, x+2
then the sum of their squares = x² + (x+1)² + (x+2)²
given that the sum =365
∴x² + (x+1)² + (x+2)² = 365
⇒x² + x² + 2x + 1 + x² + 2* 2* x + 2² = 365
⇒3x² + 2x + 1 + 4x + 4 = 365
⇒3x² + 6x + 5 = 365
⇒3x² + 6x +5 - 365 = 0
⇒3x² + 6x - 360 = 0
⇒3[ x² + 2x - 120] = 0
⇒x² + 2x - 120 = 0
⇒x² + 12x - 10x - 120 =0
⇒x(x + 12) - 10(x + 12) = 0
⇒(x + 12) (x - 10) = 0
⇒x+12 = 0   or   x-10 = 0
⇒x = -12    or x = 10
⇒x can not be -12 as the consecutive no`s are positve and -12 is negative. 
∴x = 10 
∴the three consecutive no`s are x= 10
x+1 = 11, x+2 = 12
10, 11, 12 are the no`s