# Find two consecutive positive integers, sum whose squares is 365.

2
by giaseerat

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.
• 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 !!