Last updated at May 29, 2018 by Teachoo

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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 & sum of roots = (โ๐)/๐ 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 & sum of roots = (โ๐)/๐ 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, โฆ. 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 + ๐)/(๐ โ ๐) = 17/15 Taking L.H.S (๐ + ๐)/(๐ โ ๐) Putting value of p = ab & q = cd from (2) & (4) = (๐๐ + ๐๐)/(๐๐ โ ๐๐) We know that a, ar , ar2 , ar3, โฆ. 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 + ๐)/(๐ โ ๐) = 17/15 Taking L.H.S (๐ + ๐)/(๐ โ ๐) Putting value of p = ab & q = cd from (2) & (4) = (๐๐ + ๐๐)/(๐๐ โ ๐๐) We know that a, ar , ar2 , ar3, โฆ. 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 + ๐)/(๐ โ ๐) = 17/15 Taking L.H.S (๐ + ๐)/(๐ โ ๐) Putting value of p = ab & q = cd from (2) & (4) = (๐๐ + ๐๐)/(๐๐ โ ๐๐) = (๐๐ + ๐๐)/(๐๐ โ ๐๐) Putting values b = ar , c = ar2 , d = ar3 = ((๐๐^2 )(๐๐^3 ) + ๐(๐๐))/((๐๐^2 )(๐๐^3 ) โ ๐(๐๐)) = (๐2๐4 + ๐2๐)/(๐2๐4 โ ๐2๐) = (๐2๐4 + ๐2๐)/(๐2๐4 โ ๐2๐) = (๐2๐(๐4 + 1))/(๐2๐(๐4 โ 1 )) = (๐4 + 1 )/(๐4 โ 1) So, (๐ + ๐)/(๐ โ ๐) = (๐4 + 1 )/(๐4 โ 1), we need to find r first. = (๐๐ + ๐๐)/(๐๐ โ ๐๐) Putting values b = ar , c = ar2 , d = ar3 = ((๐๐^2 )(๐๐^3 ) + ๐(๐๐))/((๐๐^2 )(๐๐^3 ) โ ๐(๐๐)) = (๐2๐4 + ๐2๐)/(๐2๐4 โ ๐2๐) = (๐2๐4 + ๐2๐)/(๐2๐4 โ ๐2๐) = (๐2๐(๐4 + 1))/(๐2๐(๐4 โ 1 )) = (๐4 + 1 )/(๐4 โ 1) So, (๐ + ๐)/(๐ โ ๐) = (๐4 + 1 )/(๐4 โ 1), we need to find r first. Now Dividing (1) & (3) (๐ + ๐)/(๐ + ๐) = 3/12 Putting values b = ar , c = ar2 , d = ar3 (๐ + ๐๐)/(๐๐2 +๐๐3) = 3/12 (๐(1 + ๐))/(๐๐2(1 + ๐)) = 3/12 1/๐2 = 3/12 1/๐2 = 1/4 r2 = 4 Now, (๐ + ๐)/(๐ โ ๐) = (๐4 + 1 )/(๐4 โ 1), Putting r2 = 4 = (4^2 + 1 )/(4^2 โ 1) = (16 + 1)/(16 โ 1) = (17 )/15 = R.H.S Thus, L.H.S = R.H.S Hence proved

Miscellaneous

Misc 1

Misc 2

Misc 3 Important

Misc 4

Misc 5

Misc 6 Important

Misc 7 Important

Misc 8

Misc 9

Misc 10 Important

Misc 11

Misc 12

Misc 13

Misc 14 Important

Misc 15

Misc 16 Important

Misc 17

Misc 18 You are here

Misc 19 Important

Misc 20

Misc 21 (i) Important

Misc 21 (ii)

Misc 22 Important

Misc 23 Important

Misc 24 Deleted for CBSE Board 2022 Exams

Misc 25 Important Deleted for CBSE Board 2022 Exams

Misc 26 Deleted for CBSE Board 2022 Exams

Misc 27

Misc 28 Important

Misc 29 Important

Misc 30

Misc 31 Important

Misc 32 Important

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.