 # if a,b,c are in g.p. , then the equation ax2 + bx + c = 0 & dx2+2ex+f=0 have a common root if d/a, e/b, f/c are in(a) A.P.(b)G.P.(c)H.P.(d)NONE OF THESE SHAIK AASIF AHAMED
7 years ago
Hello student,
a, b, c are in GP
⇒b2=ac ------------(1)
Now,ax2+2bx+c=0
⇒x=−2b±sqrt(4b2−4ac)/2a
⇒x=(−b±0)/a[Using(1)]
⇒x=−b/a
Also,dx2+2ex+f=0
⇒x=−2e±sqrt(4e2−4df)/2d
⇒x=−e+sqrt(e2−df)/d
Now, since the 2 equations have 1 common root, thus:-
−b/a=−e+sqrt(e2−df)/d,
⇒d/a=(e/b)−sqrt(e2−df)/b,
⇒d/a=e/b−sqrt((e2−df)/b2)
⇒(e2−df)/b2=(e/b−d/a)2,
⇒e2/b2−df/ac=(e/b−d/a)2, [Using(1)]
⇒(e/b)2−(d/a)(f/c)=(e/b)2−2(e/b)(d/a)+(d/a)2,
⇒−(da)(fc)=−2(e/b)(d/a)+(d/a)2,
⇒f/c+d/a=2(e/b),
⇒d/a,e/b,f/careinAP
Hence a is the correct answer Kushagra Madhukar
2 years ago
Dear student,

a, b, c are in GP
⇒ b2 = ac ------------(1)
Now, ax2+2bx+c=0
⇒x=−2b±sqrt(4b2−4ac)/2a
⇒x=(−b±0)/a [Using(1)]
⇒x=−b/a
Also,dx2+2ex+f=0
⇒x=−2e±sqrt(4e2−4df)/2d
⇒x=−e+sqrt(e2−df)/d
Now, since the 2 equations have 1 common root, thus:-
−b/a=−e+sqrt(e2−df)/d,
⇒d/a=(e/b)−sqrt(e2−df)/b,
⇒d/a=e/b−sqrt((e2−df)/b2)
⇒(e2−df)/b2=(e/b−d/a)2,
⇒e2/b2−df/ac=(e/b−d/a)2, [Using(1)]
⇒(e/b)2−(d/a)(f/c)=(e/b)2−2(e/b)(d/a)+(d/a)2,
⇒−(da)(fc)=−2(e/b)(d/a)+(d/a)2,
⇒f/c+d/a=2(e/b),
⇒d/a,e/b,f/c are in AP.
Hence a is the correct answer

Hope it helps.
Thanks and regards,
Kushagra