# Example 11

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 11 Find two consecutive odd positive integers, sum of whose squares is 290. There is a difference of 2 in odd consecutive integers, Hence, Let the first integer = x Second integer = x + 2 Also, given that Sum of the squares of both the numbers = 290 (First number)2 + (Second number) 2 = 290 (x)2 + (x + 2)2 = 290 Using (a + b)2 = a2 + b2 + 2ab x2 + x2 + 4 + 4x = 290 2x2 + 4x + 4 – 290 = 0 2x2 + 4x – 286 = 0 Divide the equation by 2 2𝑥2/2+4𝑥/2−286/2= 0/2 x2 + 2x – 143 = 0 We solve by quadratic formula Comparing equation with ax2 + bx + c = 0 Here, a = 1, b = 2, c = – 143 We know that, D = b2 – 4ac D = (2)2 – 4 ×1×(−143) D = 4 + (4 ×143) D = 4 + 572 D = 576 Hence, roots to equation are given by x = (− 𝑏 ± √𝐷)/2𝑎 Putting values x = (− 2 ± √576)/(2 × 1) x = (− 2 ± √576)/(2 ) x = (− 2 ± 24)/2 Solving So, x = 11 & x = – 13 But x is a positive odd integer So, x = 11 is the solution ∴ First odd integer = x = 11 & Second consecutive integer = x + 2 = 11 + 2 = 13

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .