Question

what are the roots of the equation (x^3+4x^2+6x+12)=0 how am i supposed to find this ?? help....pleaseeee

Answer 1

any similar type of sum done before ?

Answer 2

do u know about hit and trail method

Answer 3

roots of this are horrible....can't apply hit and trail...
is the question correct ?@erica.d

Answer 4

no this is the first one...and i haven't done this before...i don't know about hit and trial :(

Answer 5

do i go for calculator?

Answer 6

For a cubic equation like:
Ax^3 + Bx^2 + Cx + D = 0,
having three roots as a, b, c; the relations between these roots are given by:
a*b*c = -D/A
a+b+c = -B/A
ab+bc+ca = C/A

Answer 7

that's horrible

Answer 8

Have you ever tried solving Quad roots ?

Answer 9

yes...but this is cubic

Answer 10

I know I jus Wanted to Know Because Solving Cubic Roots Is Diff

Answer 11

@satellite73 .....

Answer 12

So, for the equation,
x^3+4x^2+6x+12=0
A=1,B=4,C=6,D=12
The product of the roots is:
abc= -D/A = -12/1=-12

Answer 13

The sum of the roots is :
a+b+c= -B/A=-4/1=-4

Answer 14

i would like to see how @satellite73 approaches this............

Answer 15

@satellite73 pleaseeeeee @satellite73 and thank you @hba

Answer 16

@cwrw238

Answer 17

@ganeshie8

Answer 18

I am Pretty Sure I am correct :)

Answer 19

i guess i'll do it in a while :)

Answer 20

@mukushla

Answer 21

analyticall approach will be painfull

Answer 22

LoL @mukushla

Answer 23

lol

Answer 24

hba ur method also will be back to original equation...

Answer 25

why you guys are laughing ?? :(

Answer 26

nothing

we can change the equation to the form\[x^3+px+q=0\]and this one is solved by hartnn
where is ur toturial hartnn?

Answer 27

I know that :)

Answer 28

http://openstudy.com/study#/updates/504335bee4b000724d4620b4

Answer 29

but how do we change ?

Answer 30

for \[ax^3+bx^2+cx+d=0\]\[t=x+\frac{b}{3a}\]will change the equation to\[x^3+px+q=0\]if im not wrong

Answer 31

but numerical methods like newton-raphson is better for such equations

Answer 32

here's is a method... find the derivative of the function. You can see that is always positive.
So, the function is monotonically increasing. So it will have only 1 real root.
The other two roots are complex.
So the roots are r, p+iq and p-iq
Put these in the equations given by @hba and you have 3 equations and 3 variables.
Solve for r,p and q.

Answer 33

here is a nice article
http://en.wikipedia.org/wiki/Cubic_function

Answer 34

as already told, maybe solving those will give back u original question...

Answer 35

let me check...

Answer 36

hmmm... you're right :\

Answer 37

i think when we encounter cubic equations with integer coefficients better to go with rational root theorem directly...and checking for possible rational roots. if there is not any id say just using numerical methods

Answer 38

otherwise if u want solve it analyticall it will be painfull :)