Question

please answer fast , find two consecutive positive integers, sum of whose squares is 365?

Answer 1

let the consecutive no be x and x+1

a/q
x^2+(x+1)^2=365
x^2+x^2+2x+1-365=0
2x^2+2x-364=0
2x^2+28x-26x-364=0
2x(x+14)-26(x+14)=0
(2x-26) (x+14) =0
either. or
2x=26. x=-14
x=26/2
x=13

since nos are positive ,we will take value of x=13

Now. x+1=13+1=14

Answer 2

Heya user !!

Here's the answer you are looking for

Let the 2 consecutive numbers be x and (x + 1).

The question says,

x² + (x + 1)² = 365

x² + x² + 2x + 1 = 365

2x² + 2x - 364 = 0

x² + x - 182 = 0

x² - 13x + 14x - 182 = 0

(x - 13)(x + 14) = 0

➡️Thus, x = 13 or x = –14

Since, x is a positive integer, x = 13 is our answer.

Therefore, the 2 consecutive positive integers, sum of whose square is 364 are 13 and 14.


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