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Example 11 - Find two consecutive odd positive integers - Examples

  1. Chapter 4 Class 10 Quadratic Equations
  2. Serial order wise
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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 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

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Davneet Singh
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