Given below are three equations i.e. 'I', 'II' and 'III'. If roots of the equation 'I' are 'p' and 'q', respectively such that p > q while the roots of the equation 'II' are 'm' and 'n' respectively such that m > n, then find the value of 'K'. I. 2x²- 6(x + 4) = 3x + 11 II. 2y²- 12(y - 4) = 7y + 6 III. (15m/2) - 8q = 9p - 4n + K²

Explanation

2 Detailed solution: From I: 2x²- 6(x + 4) = 3x + 11 Or, 2x²- 6x - 24 - 3x - 11 = 0 Or, 2x²- 9x - 35 = 0 Or, 2x²- 14x + 5x - 35 = 0 Or, 2x(x - 7) + 5(x - 7) = 0 Or, (2x + 5)(x - 7) = 0 So, x = -2.5 or 7 So, p = 7 and q = -2.5 From II: 2y²- 12(y - 4) = 7y + 6 Or, 2y²- 12y + 48 - 7y - 6 = 0 Or, 2y²- 19y + 42 = 0 Or, 2y²- 12y - 7y + 42 = 0 Or, 2y(y - 6) - 7(y - 6) = 0 Or, (2y - 7)(y - 6) = 0 So, y = 3.5 or 6 So, m = 6 and n = 3.5 From III: (15m/2) - 8q = 9p - 4n + K² Or, {(15 X 6)/2} - 8 X (-2.5) = 9 X 7 - 4 X 3.5 + K² Or, 45 + 20 = 63 - 14 + K² Or, 65 = 49 + K² So, K = ±4