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Ex 4.2, 4 - Find two consecutive positive integers - Ex 4.2

Ex 4.2, 4 - Chapter 4 Class 10 Quadratic Equations - Part 2

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Ex 4.2 ,4 Find two consecutive positive integers, sum of whose squares is 365. There is difference of 1 in consecutive positive integers Let First integer = x So, Second integer = x + 1 Also given that Sum of squares = 365 (First number)2 + (Second number)2 = 365 x2 + (x + 1)2 = 365 x2 + x2 + 12 + 2 × x ×1=365 2x2 + 1 + 2x = 365 2x2 + 2x +1 – 365 = 0 2x2 + 2x – 364 = 0 2 (x2 + x – 182) = 0 x2 + x – 182 = 0/2 x2 + x – 182 = 0 We factorize by splitting the middle term method x2 + 14x – 13 x – 182 = 0 x (x + 14) – 13 (x + 14) = 0 (x – 13) (x + 14) = 0 So the root of the equation are x = 13 & x = – 14 Since we have to find consecutive positive numbers We take x = 13 First number = x = 13 Second number = x + 1 = 13 + 1 = 14

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.