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

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20,21

13,14

15,16

25,26

answer is B.

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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: