Example 23

Transcript

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 𝑏/𝑎 = 𝑐/𝑏 = 𝑑/𝑐 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 𝑏/𝑎 = 𝑐/𝑏 = 𝑑/𝑐 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 𝑏/𝑎 = p Solving (bp – c) = 0 bp = c 𝑐/𝑏 = p Solving (cp – d) = 0 cp = d 𝑑/𝑐 = p This implies that (b )/a = (c )/b = (d )/c = p Hence a, b, c and d are in G.P.