Misc 18 - If a, b are roots of x2 - 3x + p = 0, c,d are roots - Geometric Progression(GP): Calculation based/Proofs

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Misc 18 If a and b are the roots of x2 โ€“ 3x + p = 0 and c,d are roots of x2 โ€“ 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q โ€“ p) = 17:15. Introduction For quadratic equation ax2 + bx + c = 0 Product of roots = ๐‘/๐‘Ž & sum of roots = (โˆ’๐‘)/๐‘Ž Misc 18 If a and b are the roots of x2 โ€“ 3x + p = 0 and c,d are roots of x2 โ€“ 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q โ€“ p) = 17:15. Introduction For quadratic equation ax2 + bx + c = 0 Product of roots = ๐‘/๐‘Ž & sum of roots = (โˆ’๐‘)/๐‘Ž Misc 18 If a and b are the roots of x2 โ€“ 3x + p = 0 and c,d are roots of x2 โ€“ 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q โ€“ p) = 17:15. We know that a, ar , ar2 , ar3, โ€ฆ. are in G.P. with first term a & common ratio r Given a, b, c, d are in G.P. So, a = a b = ar c = ar2 d = ar3 We have to prove (๐‘ž + ๐‘)/(๐‘ž โˆ’ ๐‘) = 17/15 Taking L.H.S (๐‘ž + ๐‘)/(๐‘ž โˆ’ ๐‘) Putting value of p = ab & q = cd from (2) & (4) = (๐‘๐‘‘ + ๐‘Ž๐‘)/(๐‘๐‘‘ โˆ’ ๐‘Ž๐‘) We know that a, ar , ar2 , ar3, โ€ฆ. are in G.P. with first term a & common ratio r Given a, b, c, d are in G.P. So, a = a b = ar c = ar2 d = ar3 We have to prove (๐‘ž + ๐‘)/(๐‘ž โˆ’ ๐‘) = 17/15 Taking L.H.S (๐‘ž + ๐‘)/(๐‘ž โˆ’ ๐‘) Putting value of p = ab & q = cd from (2) & (4) = (๐‘๐‘‘ + ๐‘Ž๐‘)/(๐‘๐‘‘ โˆ’ ๐‘Ž๐‘) We know that a, ar , ar2 , ar3, โ€ฆ. are in G.P. with first term a & common ratio r Given a, b, c, d are in G.P. So, a = a b = ar c = ar2 d = ar3 We have to prove (๐‘ž + ๐‘)/(๐‘ž โˆ’ ๐‘) = 17/15 Taking L.H.S (๐‘ž + ๐‘)/(๐‘ž โˆ’ ๐‘) Putting value of p = ab & q = cd from (2) & (4) = (๐‘๐‘‘ + ๐‘Ž๐‘)/(๐‘๐‘‘ โˆ’ ๐‘Ž๐‘) = (๐‘๐‘‘ + ๐‘Ž๐‘)/(๐‘๐‘‘ โˆ’ ๐‘Ž๐‘) Putting values b = ar , c = ar2 , d = ar3 = ((๐‘Ž๐‘Ÿ^2 )(๐‘Ž๐‘Ÿ^3 ) + ๐‘Ž(๐‘Ž๐‘Ÿ))/((๐‘Ž๐‘Ÿ^2 )(๐‘Ž๐‘Ÿ^3 ) โˆ’ ๐‘Ž(๐‘Ž๐‘Ÿ)) = (๐‘Ž2๐‘Ÿ4 + ๐‘Ž2๐‘Ÿ)/(๐‘Ž2๐‘Ÿ4 โˆ’ ๐‘Ž2๐‘Ÿ) = (๐‘Ž2๐‘Ÿ4 + ๐‘Ž2๐‘Ÿ)/(๐‘Ž2๐‘Ÿ4 โˆ’ ๐‘Ž2๐‘Ÿ) = (๐‘Ž2๐‘Ÿ(๐‘Ÿ4 + 1))/(๐‘Ž2๐‘Ÿ(๐‘Ÿ4 โˆ’ 1 )) = (๐‘Ÿ4 + 1 )/(๐‘Ÿ4 โˆ’ 1) So, (๐‘ž + ๐‘)/(๐‘ž โˆ’ ๐‘) = (๐‘Ÿ4 + 1 )/(๐‘Ÿ4 โˆ’ 1), we need to find r first. = (๐‘๐‘‘ + ๐‘Ž๐‘)/(๐‘๐‘‘ โˆ’ ๐‘Ž๐‘) Putting values b = ar , c = ar2 , d = ar3 = ((๐‘Ž๐‘Ÿ^2 )(๐‘Ž๐‘Ÿ^3 ) + ๐‘Ž(๐‘Ž๐‘Ÿ))/((๐‘Ž๐‘Ÿ^2 )(๐‘Ž๐‘Ÿ^3 ) โˆ’ ๐‘Ž(๐‘Ž๐‘Ÿ)) = (๐‘Ž2๐‘Ÿ4 + ๐‘Ž2๐‘Ÿ)/(๐‘Ž2๐‘Ÿ4 โˆ’ ๐‘Ž2๐‘Ÿ) = (๐‘Ž2๐‘Ÿ4 + ๐‘Ž2๐‘Ÿ)/(๐‘Ž2๐‘Ÿ4 โˆ’ ๐‘Ž2๐‘Ÿ) = (๐‘Ž2๐‘Ÿ(๐‘Ÿ4 + 1))/(๐‘Ž2๐‘Ÿ(๐‘Ÿ4 โˆ’ 1 )) = (๐‘Ÿ4 + 1 )/(๐‘Ÿ4 โˆ’ 1) So, (๐‘ž + ๐‘)/(๐‘ž โˆ’ ๐‘) = (๐‘Ÿ4 + 1 )/(๐‘Ÿ4 โˆ’ 1), we need to find r first. Now Dividing (1) & (3) (๐‘Ž + ๐‘)/(๐‘ + ๐‘‘) = 3/12 Putting values b = ar , c = ar2 , d = ar3 (๐‘Ž + ๐‘Ž๐‘Ÿ)/(๐‘Ž๐‘Ÿ2 +๐‘Ž๐‘Ÿ3) = 3/12 (๐‘Ž(1 + ๐‘Ÿ))/(๐‘Ž๐‘Ÿ2(1 + ๐‘Ÿ)) = 3/12 1/๐‘Ÿ2 = 3/12 1/๐‘Ÿ2 = 1/4 r2 = 4 Now, (๐‘ž + ๐‘)/(๐‘ž โˆ’ ๐‘) = (๐‘Ÿ4 + 1 )/(๐‘Ÿ4 โˆ’ 1), Putting r2 = 4 = (4^2 + 1 )/(4^2 โˆ’ 1) = (16 + 1)/(16 โˆ’ 1) = (17 )/15 = R.H.S Thus, L.H.S = R.H.S Hence proved

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.