Question

Find the consecutive positive integers sum of whose squares is 365

Answer 1

Let first number = x

Then second number will one more so that next number wil x+1

Given that sum of whose squares is 365.

      x^2+ (  x + 1)^2 = 365

use formula of (a +b)^2 = a^2 + 2ab +b^2

x^2 + x^2 + 2x + 1 = 365

    x^2 +   x^2 + 2x + 1 – 365 = 0

  2x^2 + 2x – 364 =0

I take 2 common.

    x^2 +  x – 182 = 0. factorize it now

x^2 + 14x - 13x - 182 = 0

x(x + 14) - 13(x + 14) = 0

(x + 14) (x - 13) = 0

x + 14 = 0 or x - 13 = 0

x = -14 or x = 13

The number can be positive because in question it required consecutive positive integers.

Thus x = 13

and x + 1 = 13 + 1 = 14

#### I hope so you like my answer ####

Answer 2

Let, First number be x and

Second number be (x+1).

eqn will be:

x² +(x+1)² = 365

x² +x²+2x+1²=365

2x²+2x+1 =365

2x²+2x =364

x²+x =182

(x+1/2)² =182+(1/2)²

x+ 1/2 =√729/√4

x =± 27/2 - 1/2

x=13 or -14

Given , the consecutive numbers are +ve.

The numbers are therefore, 13 and 14

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