Question

find two consecutive positive integers, sum whose squares is 365.

Answer 1

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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Answer 2

Answer:

13 and 14 are the two consecutive positive integers , sum of whose square is 365 .

Step-by-step explanation:

Explanation:

Let two consecutive positive integers be x and (x+ 1)

Now ,according to question , two consecutive positive integers, sum  of whose square is 365 .

⇒ $x^2 + (x+1) ^2 = 365$

Step 1:

On solving the equation $x^2 + (x+1) ^2 = 365$ we get ,

⇒ $x^2 + x^2 + 2x + 1 = 365$

⇒2$x^2 + 2x + 1 = 365$

⇒$2x^2 + 2x - 364$ = 0

Taking  2  as common  ,

$x^2 + x- 182 = 0$

Now by middle term splitting method ,

⇒$x^2 + 14x - 13x - 182 = 0$

⇒x (x+ 14) - 13 (x+ 14 ) = 0

⇒(x+ 14 ) (x- 13 ) = 0

⇒ x = -14 and x = 13

Step 2:

But given in the question that  integers are positive. so , we take  the value of x = 13 .

So , one  of the integer  is 13 and the other is  (x+ 1) = 13 + 1= 14 .

Final answer:

Hence , 13 and 14 are the two consecutive positive integers .

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