Question

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

Answer 1

Answer:

  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 .

Step-by-step explanation:

Answer 2

Answer :-

Let the consecutive positive integers be x and x +1

$\sf{ \bold{Given \: That \: x^{2} + (x + 1) ^{2} = 365}}$

$\sf{ \bold{\implies \: x^{2} + {x}^{2} + 1 + 2x = 365}}$

$\sf{ \bold{ \implies \: 2 {x}^{2} + 2x - 364 = 0}}$

$\sf{ \bold{ \implies \: {x}^{2} + x - 182 = 0}}$

$\sf{ \bold{\implies \: {x}^{2} + 14x - 13x - 182 = 0}}$

$\sf{ \bold{\implies \: x(x + 14) - 13(x + 14) = 0}}$

$\sf{ \bold{ \implies \: (x + 14)(x - 13) = 0}}$

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

$\sf{ \bold{ \color{blue}{ \therefore \: x + 1 = 13 + 1 = 14}}}$

Therefore, two consecutive positive integers will be 13 and 14