Question

if the roots of the equation ax^2 + bx + c =0 are of the form x/(x-1) and (x+1)/x then the value of (a+b+c)^2 is?

Answer 1

\[(a + b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\]

Will this help or not?

Answer 2

@experimentX ,,product of roots is not 1..maybe your mis-read..

Answer 3

Oh .. i misread.

Answer 4

product of roots = (x+1)/(x-1) = c/a
sum of roots = (2x^2 - 1)(x^2 -1) = -b/a
=> (c+b)/a = (x+1)/(x-1) - (2x^2 - 1)(x^2 -1) = something..
i dont know how to eliminate a now..

Answer 5

as c+b = f(a)

Answer 6

Sum Of Roots: -b/a

And product of Roots = c/a

Answer 7

\[r1*r2=\frac{x+1}{x-1}\]
\[\frac{c}{a}=\frac{x+1}{x-1}\]
use componendo and dividendo to find x.....

Answer 8

seems we can add it too ..

Answer 9

\[\frac{x}{x-1} + \frac{x+1}{x} = \frac{-b}{a}\]

Answer 10

\[= \frac{x^2 + x^2 - 1}{x(x-1)} = \frac{2x^2 - 1}{x(x-1)} = \frac{-b}{a}\]

Answer 11

we still need one equation.

Answer 12

i got it..i min i'll post it

Answer 13

i mean 1 min**

Answer 14

product of roots = (x+1)/(x-1) = c/a
=> x= (c+a)/(c-a)
on substitutions,
we see roots are :
(c+a)/2a and 2c/(a+c)
now sum of roots = -b/a = [(a+c)^2 + 4ac ]/(2a(a+c) = -b/a
=> a^2 + c^2 +6ac = -2ab -2bc
=>(a+b+c)^2 = b^2 - 4ac
= discriminant ..
oops,,maybe you want it in terms of x only ?

Answer 15

can you tell me the ans if you have it @Yahoo!

Answer 16

b^2-4ac

Answer 17

i mean the book's ans ?

Answer 18

yup man u r correct!!!

Answer 19

hmm,,phew!! :P