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Example 10 - 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
Example 8 Important
Example 9 Important
Example 10 You are here
Example 11
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 10 Solve Q. 2 (i) of Exercise 4.1 by using the quadratic formula. The equation we formed in Ex4.1,2 (i) was 2x2 + x 528 = 0 Comparing equation with ax2 + bx + c = 0 Here a = 2, b = 1, c = 528 We know that D = b2 4ac D = (1)2 4 2 ( 528) D = 1 8 ( 528) D = 1 + 8 528 D = 1 + 4224 D = 4225 So, the roots of the equation are given by. x = ( )/2 Putting values x = ( 1 4225)/(2 2) x = ( 1 4225)/4 x = ( 1 65)/4 Solving Hence, x = 16 and x = ( 33)/2 But x cannot be negative as number of breadth cannot be negative So, Breadth of the hall = x = 16 m & Length of the hall = 2x + 1 = 2 (16) + 1 = 33 metre .
