Question 4 - Examples - Chapter 4 Class 10 Quadratic Equations

Last updated at Dec. 13, 2024 by

Example 10 - Solve Q. 2 (i) of Ex 4.1 by quadratic formula - Examples

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

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Question 4 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 .

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo