Question

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

Answer 1

Answer:

HOPE THIS HELPS YOU

PLZZ MARK AS BRAINLIEST

Answer 2

Answer:

Step-by-step explanation:

let the no. be  x and x+1

 then we have x²+(x+1)²=365

                    2x²+2x-364=0

                    x²+x-182=0

                    x²+14x-13x-182=0

                    (x+14)(x-13)=0

                That gives x=13 Avoiding negative value

                one number is 13 and other one 14.

or

Solution:

      Let the two consecutive Numbers be x and x+1.

  Therefore ,

          x² + (x+1)² = 365

          x² + x² +1² + 2*x*1 = 365 (because (A+B)² = A² + B²+ 2AB)

      or 2x² + 1 + 2x = 365

        2x² + 2x = 365 - 1

        2x² + 2x = 364

        2(x² + x) = 364

      or x² + x = 364/2

      or x² + x = 182

    or x² + x - 182 =0

    Now Solve the Quadratic Equation ,

            x² + 14x - 13x - 182 = 0

  Note : - 13 *14 = 182 , this is because I write 14x - 13x instead of x , so as to solve the quadratic equation .

          x (x+14) - 13 (x +14 ) = 0

         ( x- 13 )(x+14)=0

        Therefore , Either x - 13 = 0 or x+14 =0

            Since the Consecutive Integers are positive ,

          therefore , x-13 = 0

                  ⇒ x =13

hence One of the Positive Integers = 13 ,

therefore other positive integer = x+1 = 13+1 = 14

So the two consecutive positive Integers are 13 and 14 .

Hope this helps You !!

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