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Example 23 - If a, b, c, d, p are different real numbers - Examples

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Example 23, If a, b, c, d and p are different real numbers such that (a2 + b2 + c2)p2 2(ab + bc + cd) p + (b2 + c2 + d2) 0, then show that a, b, c and d are in G.P. Introduction If x2 + y2 + z2 0 So, if x2 + y2 + z2 0, x2 + y2 + z2 = 0 i.e. x = 0, y = 0 , z = 0 Example 23, If a, b, c, d and p are different real numbers such that (a2 + b2 + c2)p2 2(ab + bc + cd) p + (b2 + c2 + d2) 0, then show that a, b, c and d are in G.P. We need to show a, b, c & d are in GP So, common ratio must be same we have to prove / = / = / It is given that (a2 + b2 + c2) p2 2 (ab + bc + cd) p + (b2 + c2 + d2) 0 Solving a2p2 + b2p2 + c2p2 2abp 2bcp 2cdp + b2 + c2 + d2 0 (ap)2 + b2 2abp + (bp)2 + c2 2bcp + (cp)2 + d2 2cdp 0 Example 23, If a, b, c, d and p are different real numbers such that (a2 + b2 + c2)p2 2(ab + bc + cd) p + (b2 + c2 + d2) 0, then show that a, b, c and d are in G.P. We need to show a, b, c & d are in GP So, common ratio must be same we have to prove / = / = / It is given that (a2 + b2 + c2) p2 2 (ab + bc + cd) p + (b2 + c2 + d2) 0 Solving a2p2 + b2p2 + c2p2 2abp 2bcp 2cdp + b2 + c2 + d2 0 (ap)2 + b2 2abp + (bp)2 + c2 2bcp + (cp)2 + d2 2cdp 0 Solving (ap b) = 0 ap = b / = p Solving (bp c) = 0 bp = c / = p Solving (cp d) = 0 cp = d / = p This implies that (b )/a = (c )/b = (d )/c = p Hence a, b, c and d are in G.P.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.