Chapter 4 Class 10 Quadratic Equations (Term 2)

Serial order wise

Last updated at Aug. 3, 2021 by Teachoo

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