(p + q) = 8 and (p2 + q2 - 6) =

Solution

ATQ, p + q = 8 -------- (I) And, p² + q\² = 34 We know that, (p + q)² = p² + q² + 2pq So, (8)² = 34 + 2pq Or, pq = {(64 – 34)/2} So, pq = 15 Or, q = (15/a) On substituting value of ‘q’ in equation (I), we have; p + (15/p) = 8 Or, p² + 15 = 8p Or, p² – 8p + 15 = 0 Or, p² – 5p – 3p + 15 = 0 Or, p² – 5p – 3p + 15 = 0 Or, p(p – 5) – 3(p – 5) = 0 Or, (p – 3) (p – 5) = 0 So, p = 3 or 5 By putting the value of ‘p’ in equation (I), we get q = 5 or 3 Since, p < q.  So, p = 3 and q = 5 therefore, required value = (p/q) = (3/5)