## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

##### Q. No. 4 of Exercise 10E

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4
##### The sum of the squares of two consecutive positive integers is 365. Find the integers.

Let the required two consecutive positive integers be x and x + 1

According to given condition,

x2 + (x + 1)2 = 365

x2 + x2 + 2x + 1 = 365 using (a + b)2 = a2 + 2ab + b2

2x2 + 2x – 364 = 0

x2 + x – 182 = 0

Using the splitting middle term – the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1 b = 1 c = – 182

= 1. – 182 = – 182

And either of their sum or difference = b

= 1

Thus the two terms are 14 and – 13

Difference = 14 – 13 = 1

Product = 14. – 13 = – 182

x2 + x – 182 = 0

x2 + 14x – 13x – 182 = 0

x(x + 14) – 13(x + 14) = 0

(x + 14) (x – 13) = 0

(x + 14) = 0 or (x – 13) = 0

x = – 14 or x = 13

x = 13 (x is a positive integer)

x + 1 = 13 + 1 = 14

Thus the required two consecutive positive integers are 13, 14

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