Last updated at May 29, 2018 by
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 / = / q2 = pr Solving the equation px2 + 2qx + r = 0 For ax2 + bx + c roots are x = ( ( 2 4 ))/2 Here a = p, b = 2q & c = r Hence the roots of equation px2 + 2qx + r = 0 are x = ( 2q (4q2 4rp))/2p Putting q2 = pr from (1) x = ( 2q (4q2 4rp))/2p = ( 2q (4pr 4rp))/2p = ( 2 0)/2 = ( 2 )/2 = ( q )/p Thus x = ( q )/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, ( q )/p is a root of dx2 + 2ex + f = 0 Putting x = ( q )/p in dx2 + 2ex + f = 0 d (( q )/p)^2 + 2e (( q )/p) + f = 0 d (( q)^2 )/p2 ((2eq )/p) + f = 0 ( ^2 )/p2 ((2eq )/p) + f = 0 ( 2 2 + 2)/ 2 = 0 dq2 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 (d )/p = / / i.e. to show (e )/q + / = / + (d )/p To show :- 2 / = / + / Now, from (2) dq2 2eap + fp2 = 0 Dividing this by pq2 dq2/pq2 2epq/pq2 + fp2/pq2 = 0/pq2 (d )/p (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 (d )/p, (e )/q, (f )/r are in A.P Hence proved
Examples
Example 1 (ii)
Example 2
Example 3 Important
Example 4
Example 5
Example 6 Important
Example 7
Example 8 Important
Example 9
Example 10 Important
Example 11
Example 12 Important
Example 13
Example 14 Important
Example 15 Important
Example 16
Example 17 Important
Example 18 Important
Example 19 Important Deleted for CBSE Board 2022 Exams
Example 20 Deleted for CBSE Board 2022 Exams
Example 21 Important
Example 22
Example 23
Example 24 Important You are here
Examples
About the Author