Question

Find two consecutive positive integers sum of whose squares is 365.​

Answer 1

let first integer be x

so second integer would be x+1

ATQ.

Sum of squares is 365

$( {first \: no.)}^{2} + ( {second \: number)}^{2} = 365 \\ ( {x)}^{2} + ( {x + 1)}^{2} = 365 \\ {x}^{2} + {x}^{2} + {1}^{2} + 2 \times x \times 1 = 365 \\ {2x}^{2} + 1 + 2x = 365 \\ 2 {x}^{2} + 2x + 1 - 365 = 0 \\ 2( {x}^{2} + x - 182) = 0 \\ {x}^{2} + x - 182 = \frac{0}{2} \\ {x}^{2} + x - 182 = 0$

we factorise by splitting the middle term :

${x}^{2} + 14x - 13x - 182 = 0 \\ x(x + 14) - 13(x + 14) = 0 \\ (x - 13)(x + 14) = 0 \\ x - 13 = 0 \\ x = 13 \\ x + 14 = 0 \\ x = - 13$

So the root of equation are :

X=13 and. X= - 14

since we have to find consecutive number .

we take X = 13

second = X+1

= 13+1

= 14

Answer 2

Step-by-step explanation:

hey guys ur 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 .

Hope this helps You !!

Thanks Cheers !!