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

**Solution:**

Let the first integer be x.

The next consecutive positive integer will be x + 1.

According to the given question, the sum of squares of x and x + 1 is 365. i.e.,

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

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

x^{2} + (x^{2} + 2x + 1) = 365 [ ∵ (a + b)^{2} = a^{2} + 2ab + b^{2}]

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

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

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

2(x^{2} + x - 182) = 0

x^{2} + x - 182 = 0

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

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

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

x - 13 = 0 and x + 14 = 0

x = 13 and x = - 14

**☛Try:** Solve the same equation using quadratic formula. You will still get the same roots.

The value of x cannot be negative (because it is given that the integers are positive). Thus, we ignore x = -14.

∴ x = 13 and x + 1 = 14

**☛ Check:** Class 10 Maths NCERT Solutions Chapter 4

**Video Solution:**

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

Class 10 Maths NCERT Solutions Chapter 4 Exercise 4.2 Question 4

**Summary:**

Two consecutive positive integers, the sum of whose squares is 365 are 13 and 14.

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