Question

the roots of the equation x^3-4x-12=0 are a,b,c Find the equation with roots a+b, b+c, c+a

Answer 1

One way would be to find a,b and c

Answer 2

sum of roots = a+b+c = -0/1 = 0
so u just have to find and equation whose roots are
-c,-a,-b.

Answer 3

IS IT ALWAYS TRUE @hartnn
sum of roots of a cubic equation =0

Answer 4

i guess a,b,c are not real

Answer 5

Hmm, I've found the sum and products of the roots but I don't know how to find a,b,and c
and @hartnn, how do I find an equation with roots -c,-a and -b?

Answer 6

if the equation is Ax^3+Bx^2+cx+d=0,
then sum of roots = -B/A is always true

Answer 7

oh. SINCE b=0

Answer 8

sum of roots=0

Answer 9

http://www.sosmath.com/algebra/factor/fac11/fac11.html

Answer 10

Ok then this may be a nice approach,
Since the roots are -a,-b,-c
(-x)^3 -4(-x)-12=0
-x^3+4x-12=0

Answer 11

lets see, u have roots as, -a,-b-,c
its product = -abc = -(-12) =12
sum = (-a-b-c) = -(a+b+c) = 0
so its of the form :
x^3+px+12 =0

Answer 12

or what @sauravshakya did.

Answer 13

u are correct, the final equation then becomes.
x^3-4x+12=0

Answer 14

YEP

Answer 15

ok, I got the rest of it but why did the roots have to be -b,-c, -a?

Answer 16

u know how to find sum and product of roots of the general equation :
Ax^3+Bx^2+Cx+D=0
?

Answer 17

yup!

Answer 18

sum = -B/A
product = D/A
now apply this to your equation x^3-4x-12=0
and tell me what u get sum as ?

Answer 19

sum is -b/a which is 4

Answer 20

opps I mean -0/4

Answer 21

its -0/1 = 0

Answer 22

opps again that's what I meant

Answer 23

so a+b+c=0
a+b=-c
b+c=-a
a+c=-b

Answer 24

oh I get it now! Thank you!!!

Answer 25

welcome :)
that was a good question.