Question

find two consecutive positive integers sum of whose square is 365 ?​

Answer 1

Step-by-step explanation:

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 !!

Answer 2

Answer:

$13 \: and \: 14$

Step-by-step explanation:

It is the correct answer.

Hope this attachment helps.