Question

Find two consecutive positive integers, Sum of whose square is 365.

Answer 1

Hey there,

Your answer.

-》Let the consecutive positive integers be xand x + 1.
Given that x2 +(x+1)2 = 365
= x2 + x2 +1+2x = 365
= 2x2 +2x – 364 = 0
= x2 + x – 182 = 0
= x2 + 14x – 13x – 182 = 0
= x(x+14) – 13(x+14) = 0
= (x+14)(x – 13) = 0
Either x + 14 = 0 or x − 13 = 0, i.e, x = −14 or x= 13
Since the integers are positive, x can only be 13.
∴ x + 1 = 13 + 1 = 14
Therefore, two consecutive positive integers will be 13 and 14.

:-) Hope it helps....

Answer 2

let the first no. be - x then next be - x+1
⇒ since ..it's the square of these nos.
⇒x²+ [ x+1]²=365
⇒x²+x²+2x+1=365
2x²+2x-364=0
x²+x-182=0
x²+14x-13x-182=0
therfore the 2 Consecutive nos. are 13 and 14