Find the consecutive positive integers, the sum of whose square is 365.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

20,21

13,14

15,16

25,26

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Consider the first number to be x so that the consecutive number will be x+1
Now, according to the question, the sum of the squares x and x+1 is 365. So,

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

Solve the equation

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

by using the splitting the middle term method to determine the value of x as:

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

x^{2} + x^{2} + 2 x + 1 = 365

2 x^{2} + 2 x + 1 - 365 = 0

2 x^{2} + 2 x - 364 = 0

x^{2} + x - 182 = 0

Now, split the middle term into two parts such that the sum will be the coefficient of x and the product will be the product of the coefficient of x2 and x0.

x^{2} + 14 x - 13 x - 182 = 0

x (x + 14) - 13 (x + 14) = 0

(x + 14) (x - 13) = 0

x = 13, - 14

Because the question only asks for positive consecutive numbers, disregard x=14, and the desired value of x is 13.
Now, substitute x=13 in the function x+1 so as to determine the value of the second number as: