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.,
x2 + ( x + 1)2 = 365
x2 +( x + 1)2 = 365
x2 + (x2 + 2x + 1) = 365 [ ∵ (a + b)2 = a2 + 2ab + b2]
2x2 + 2x + 1 = 365
2x2 + 2x + 1- 365 = 0
2x2 + 2x - 364 = 0
2(x2 + x - 182) = 0
x2 + x - 182 = 0
x2 + 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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