Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

if a,b,c are in g.p. , then the equation ax 2 + 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

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


 


 


 

Grade:12

2 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
6 years ago
Hello student,
Please find the answer to your question below
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
askIITians Faculty 629 Points
9 months ago
Dear student,
Please find the attached solution to your question

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

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free