Question

find the two consecutive positive integers, sum of whose squares is 365​

Answer 1

Answer:

Step-by-step explanation:

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.

     x2+ (  x + 1)2 = 365

use formula of (a +b)2 = a2 + 2ab +b2

   x2+   x2+ 2x + 1 – 365 = 0

 2x2+ 2x – 364 = 0

Divide by 2 to simplify it

   x2+   x – 182 = 0

factorize it now

Answer 2

Answer:

let one number be :- x and other number be :- x+1

                   According To Question

their sum's square:-( x+x+1)²=365

   x²+x²+1+2x=365 (by using identity a²+b²+2ab)

                 2x²+2x+1=365

       By quadratic formula:-b²-2ac

                 put the values

           (2)²-2(2)(1)=0

we get b²-2ac=0

now,x=-b±√b²-2ac÷2a

put values in this and we will get x=-1/2 or 1/2

Step-by-step explanation: