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33 votes
Find two consecutive positive integers such that the sum of their squares is 113?

User Reunanen
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1 Answer

17 votes
17 votes

7 and 8

Step-by-step explanation:

Consecutive numbers follow each other without a gap.

Let the consecutive intergers be:

x, x+1

The sum of their squares = 113

x² + (x + 1)² = 113

Expanding:

x² + (x + 1)(x + 1) = 113

x² + x² + x + x + 1 = 113

x² + x² + 2x + 1 = 113

2x²+ 2x + 1 = 113

subtract 1 from both sides:

2x² + 2x = 113 - 1

2x² + 2x = 112

divide both sides by 2:

2(x² + x) = 2(56)

x² + x = 56

x² + x - 56 = 0

Factorize the equation:

factorsof -56 whose sum gives - 1 are +8 and -7

x² + 8x - 7x - 56 = 0

x(x + 8) -7(x + 8) = 0

(x - 7)(x + 8) = 0

x - 7 = 0

x = 7

OR

x + 8 = 0

x = -8

Note: We are told the consecutive intergers are positive. This rules -8 as an answer as it is a negative interger.

when x , first consecutive number = 7

second consecutive number = x +1 = 7 + 1 = 8

Hence, the consecutive numbers are 7 and 8

User Gaurang Sondagar
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2.9k points