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Example 11 - Find two consecutive odd positive integers - Examples

Example 11 - Chapter 4 Class 10 Quadratic Equations - Part 2
Example 11 - Chapter 4 Class 10 Quadratic Equations - Part 3 Example 11 - Chapter 4 Class 10 Quadratic Equations - Part 4

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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 2 2/2+4 /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 = ( )/2 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

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Davneet Singh

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.