Question

The sum of the squares of 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 

Answer 2

Answer:

Let's consider first positive integer = x

Second positive integer = x +1

And third positive integer = x + 2

Now according to question x^2 + (x +1)^2 + (x +2)^2 = 365 => x^2 + x^2 + 1 + 2x + x^2 + 4 + 4x = 365

=> 3x^2 + 6x + 5 = 365

=> 3x^2 + 6x + 5 - 365 = 0

=> 3x^2 + 6x - 360 = 0

Or simply => x^2 + 2x - 120 = 0

=> (x + 12)(x - 10) = 0

So x = -12 and x = 10

But we know that x is positive integer. So x = -12 is not possible.

So first integer = 10

Second integer = 10 + 1 = 11

And third integer = 10 + 2 = 12                                                                                                 So the correct answer is 33                                                                                                               PLEASE MARK ME BRAINLIST