Example 11 - Chapter 4 Class 10 Quadratic Equations (Term 2)
Last updated at May 29, 2018 by Teachoo
Examples
Example 1 (i)
Important
Example 1 (ii)
Example 2 Important
Example 3
Example 4
Example 5 Important
Example 6
Example 7 Deleted for CBSE Board 2023 Exams
Example 8 Important Deleted for CBSE Board 2023 Exams
Example 9 Important Deleted for CBSE Board 2023 Exams
Example 10
Example 11 You are here
Example 12 Important
Example 13 (i)
Example 13 (ii) Important
Example 13 (iii)
Example 14 (i)
Example 14 (ii)
Example 15 Important
Example 16
Example 17 Important
Example 18
Chapter 4 Class 10 Quadratic Equations
Serial order wise
- Ex 4.1
- Ex 4.2
- Ex 4.3
- Ex 4.4
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Examples
- Case Based Questions (MCQ)
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 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
