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

Harshit Singh
3 years ago
Dear Student

Let us say, the two consecutive positive integers bexandx+ 1.

Therefore, as per the given questions,

x^2+ (x+ 1)^2= 365

⇒x^2+x^2+ 1 + 2x= 365

⇒2x^2+ 2x - 364 = 0

⇒x^2+x- 182 = 0

⇒x^2+ 14x- 13x- 182 = 0

⇒x(x+ 14) -13(x+ 14) = 0

⇒(x+ 14)(x- 13) = 0

Thus, either,x+ 14 = 0 orx- 13 = 0,

⇒x= - 14 orx= 13
since, the integers are positive, soxcan be 13, only.

x+ 1 = 13 + 1 = 14
two consecutive positive integers will be 13 and 14.

Thanks
Pankaj Suyal
13 Points
3 years ago
Let the first no. = X
Second no. = X+1 (consecutive number mean continuously)
According to the question
(X)+(X+1)= 365
X2+X2+1+2X = 365
2X2+2X=365-1
2X2+2X=364
Dividing both sides by 2
X2+X=182
X2+X-182=0
Now cwe factorize this equation
X2+14X-13X-182=0
X(X+14)-13(X+14)=0
(x+14)(x-13)=0
X+14=0
x=-14
x-13=0
x=13
Given in question positive integer so we select 13 which is value of x so can say that
X=13
X+1=13+1=14

Thank you

2X2+2X+1=365