Check sibling questions

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

Misc 18 - Chapter 9 Class 11 Sequences and Series - Part 2
Misc 18 - Chapter 9 Class 11 Sequences and Series - Part 3
Misc 18 - Chapter 9 Class 11 Sequences and Series - Part 4
Misc 18 - Chapter 9 Class 11 Sequences and Series - Part 5
Misc 18 - Chapter 9 Class 11 Sequences and Series - Part 6


Transcript

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

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.