Answers

2014-12-12T20:48:04+05:30
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.
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  • Brainly User
2014-12-13T12:57:16+05:30
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 !!

Thanks Cheers !!


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