Example 10

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 = (− 𝑏 ± √𝐷)/2𝑎 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 .