Question
I. 3p2Â - 11p + 10 = 0 II.
42q2Â + q -1 = 0 The following question contains two equations as I and II. You have to solve both equations and determine the relationship between them and give the answer.Solution
ATQ, 3p2 - 11p + 10 = 0 3p2- 6p – 5p + 10 = 0 3p(p-2) – 5(p-2) = 0 p = 5/3, 2 42q2 + q - 1 = 0 42q2 + 7q – 6q - 1 = 0 7q(6q+1) – 1(6q+1) = 0 q = 1/7 , -1/6 Hence, p > q
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