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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 – 3x + p = 0 and c,d are roots of x2 – 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q – p) = 17:15. Introduction For quadratic equation ax2 + bx + c = 0 Product of roots = 𝑐/π‘Ž & sum of roots = (βˆ’π‘)/π‘Ž Misc 18 If a and b are the roots of x2 – 3x + p = 0 and c,d are roots of x2 – 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q – p) = 17:15. Introduction For quadratic equation ax2 + bx + c = 0 Product of roots = 𝑐/π‘Ž & sum of roots = (βˆ’π‘)/π‘Ž Misc 18 If a and b are the roots of x2 – 3x + p = 0 and c,d are roots of x2 – 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q – p) = 17:15. We know that a, ar , ar2 , ar3, …. 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 (π‘ž + 𝑝)/(π‘ž βˆ’ 𝑝) = 17/15 Taking L.H.S (π‘ž + 𝑝)/(π‘ž βˆ’ 𝑝) Putting value of p = ab & q = cd from (2) & (4) = (𝑐𝑑 + π‘Žπ‘)/(𝑐𝑑 βˆ’ π‘Žπ‘) We know that a, ar , ar2 , ar3, …. 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 (π‘ž + 𝑝)/(π‘ž βˆ’ 𝑝) = 17/15 Taking L.H.S (π‘ž + 𝑝)/(π‘ž βˆ’ 𝑝) Putting value of p = ab & q = cd from (2) & (4) = (𝑐𝑑 + π‘Žπ‘)/(𝑐𝑑 βˆ’ π‘Žπ‘) We know that a, ar , ar2 , ar3, …. 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 (π‘ž + 𝑝)/(π‘ž βˆ’ 𝑝) = 17/15 Taking L.H.S (π‘ž + 𝑝)/(π‘ž βˆ’ 𝑝) Putting value of p = ab & q = cd from (2) & (4) = (𝑐𝑑 + π‘Žπ‘)/(𝑐𝑑 βˆ’ π‘Žπ‘) = (𝑐𝑑 + π‘Žπ‘)/(𝑐𝑑 βˆ’ π‘Žπ‘) Putting values b = ar , c = ar2 , d = ar3 = ((π‘Žπ‘Ÿ^2 )(π‘Žπ‘Ÿ^3 ) + π‘Ž(π‘Žπ‘Ÿ))/((π‘Žπ‘Ÿ^2 )(π‘Žπ‘Ÿ^3 ) βˆ’ π‘Ž(π‘Žπ‘Ÿ)) = (π‘Ž2π‘Ÿ4 + π‘Ž2π‘Ÿ)/(π‘Ž2π‘Ÿ4 βˆ’ π‘Ž2π‘Ÿ) = (π‘Ž2π‘Ÿ4 + π‘Ž2π‘Ÿ)/(π‘Ž2π‘Ÿ4 βˆ’ π‘Ž2π‘Ÿ) = (π‘Ž2π‘Ÿ(π‘Ÿ4 + 1))/(π‘Ž2π‘Ÿ(π‘Ÿ4 βˆ’ 1 )) = (π‘Ÿ4 + 1 )/(π‘Ÿ4 βˆ’ 1) So, (π‘ž + 𝑝)/(π‘ž βˆ’ 𝑝) = (π‘Ÿ4 + 1 )/(π‘Ÿ4 βˆ’ 1), we need to find r first. = (𝑐𝑑 + π‘Žπ‘)/(𝑐𝑑 βˆ’ π‘Žπ‘) Putting values b = ar , c = ar2 , d = ar3 = ((π‘Žπ‘Ÿ^2 )(π‘Žπ‘Ÿ^3 ) + π‘Ž(π‘Žπ‘Ÿ))/((π‘Žπ‘Ÿ^2 )(π‘Žπ‘Ÿ^3 ) βˆ’ π‘Ž(π‘Žπ‘Ÿ)) = (π‘Ž2π‘Ÿ4 + π‘Ž2π‘Ÿ)/(π‘Ž2π‘Ÿ4 βˆ’ π‘Ž2π‘Ÿ) = (π‘Ž2π‘Ÿ4 + π‘Ž2π‘Ÿ)/(π‘Ž2π‘Ÿ4 βˆ’ π‘Ž2π‘Ÿ) = (π‘Ž2π‘Ÿ(π‘Ÿ4 + 1))/(π‘Ž2π‘Ÿ(π‘Ÿ4 βˆ’ 1 )) = (π‘Ÿ4 + 1 )/(π‘Ÿ4 βˆ’ 1) So, (π‘ž + 𝑝)/(π‘ž βˆ’ 𝑝) = (π‘Ÿ4 + 1 )/(π‘Ÿ4 βˆ’ 1), we need to find r first. Now Dividing (1) & (3) (π‘Ž + 𝑏)/(𝑐 + 𝑑) = 3/12 Putting values b = ar , c = ar2 , d = ar3 (π‘Ž + π‘Žπ‘Ÿ)/(π‘Žπ‘Ÿ2 +π‘Žπ‘Ÿ3) = 3/12 (π‘Ž(1 + π‘Ÿ))/(π‘Žπ‘Ÿ2(1 + π‘Ÿ)) = 3/12 1/π‘Ÿ2 = 3/12 1/π‘Ÿ2 = 1/4 r2 = 4 Now, (π‘ž + 𝑝)/(π‘ž βˆ’ 𝑝) = (π‘Ÿ4 + 1 )/(π‘Ÿ4 βˆ’ 1), Putting r2 = 4 = (4^2 + 1 )/(4^2 βˆ’ 1) = (16 + 1)/(16 βˆ’ 1) = (17 )/15 = R.H.S Thus, L.H.S = R.H.S Hence proved

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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.