Last updated at Dec. 12, 2016 by Teachoo

Transcript

Example 24 If p,q,r are in G.P. and the equations, px2 + 2qx + r = 0 and dx2 + 2ex + f = 0 have a common root, then show that (d )/p, (e )/q, (f )/r are in A.P It is given that p, q, r are in G.P So, their common ratio is same ๐/๐ = ๐/๐ q2 = pr Solving the equation px2 + 2qx + r = 0 For ax2 + bx + c roots are x = (โ๐ ยฑโ(๐2 โ 4๐๐))/2๐ Here a = p, b = 2q & c = r Hence the roots of equation px2 + 2qx + r = 0 are x = (โ2q ยฑโ(4q2 โ4rp))/2p Putting q2 = pr from (1) x = (โ2q ยฑโ(4q2 โ4rp))/2p = (โ2q ยฑโ(4pr โ4rp))/2p = (โ2๐ยฑ0)/2๐ = (โ2๐)/2๐ = (โq )/p Thus x = (โq )/p is the root of the equation px2 + 2qx + r = 0 Also, it is given that equations px2 + 2qx + r = 0 & dx2 + 2ex + f = 0 have a common root So, (โq )/p is a root of dx2 + 2ex + f = 0 Putting x = (โq )/p in dx2 + 2ex + f = 0 d ((โq )/p)^2 + 2e ((โq )/p) + f = 0 d ((โq)^2 )/p2 โ ((2eq )/p) + f = 0 (๐๐^2 )/p2 โ ((2eq )/p) + f = 0 (๐๐2 โ2๐๐๐+๐๐2)/๐2 = 0 dq2 โ 2eap + fp2 = 0 We need to show (d )/p, (e )/q, (f )/r are in AP i.e. we need to show their common difference is some i.e. to show :- (e )/q โ (d )/p = ๐/๐ โ ๐/๐ i.e. to show (e )/q + ๐/๐ = ๐/๐ + (d )/p To show :- 2๐/๐ = ๐/๐ + ๐/๐ Now, from (2) dq2 โ 2eap + fp2 = 0 Dividing this by pq2 dq2/pq2 โ 2epq/pq2 + fp2/pq2 = 0/pq2 (d )/p โ (2e )/q + (fp )/q2 = 0 (d )/p + (fp )/q2 = (2e )/q Putting q2 = pr from (2) d/p + fp/pr = 2e/q d/p + f/r = 2e/q which is what we have to prove โด (d )/p, (e )/q, (f )/r are in A.P Hence proved

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6 Important

Example 7

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13

Example 14 Important

Example 15 Important

Example 16

Example 17 Important

Example 18 Important

Example 19 Important

Example 20

Example 21 Important

Example 22

Example 23 Important

Example 24 Important You are here

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.