Question

\[a(b-c)x^{2}+b(c-a)x+c(a-b)=0\] if the roots of given equation are equal, then a,b,c are in
a)AP
b)GP
c)HP
d)HP or GP

Answer 1

\[ (b(c-a))^2 = 4a(b−c)c(a−b)\]
Let's simplify this

Answer 2

I tried the same process.. Got some weird and really long equation.. Couldn't figure out the relation.

Answer 3

i understood only that the two equal roots are 1 each!:))

Answer 4

how do they turn out to be 1?

Answer 5

Just satisfy it. Observation :P

Answer 6

just use 1 in x's place.
the equation gets satisfied!!!!

Answer 7

a,b,c are in HP.

Answer 8

i can just give a clue:
use the relations between the roots and coefficients!

Answer 9

Exactly. :D

Answer 10

It has only only one answer and its GP.. @anusha.p you were helpful.
The roots are 1 so, \[\frac{-b(c-a)}{2a(b-c)}=1\]proceeding further i get condition for HP..

Answer 11

finally u got it:))))))

Answer 12

yeah, thanks @anusha.p , my normal approach was useless here..

Answer 13

Hold on..
It simplifies to, 2ac = ab + bc.
=> ab , ac, bc are in AP.
=> 1/c , 1/b , 1/a are in AP ( Dividing each term by abc)

Answer 14

Doesnt that mean a, b, c are in HP?

Answer 15

Yeah @siddhantsharan its in HP

Answer 16

So why'd you say, "It has only only one answer and its GP"

Answer 17

I mistyped it.. Below it i have written, "proceeding further i get condition for HP"

Answer 18

Ohhhhh. Sorry. Didnt see that. :P

Answer 19

They don't give me an option to edit answers.. And if i delete it and write it again, everything will seem too weird.. The 'flow of conversation' will be disturbed..

Answer 20

HAha. True that. Its fineee ::)