VERY INTERESTING QUESTION!!!! Find the condition that the equations ax^3+bx+c=0, and cx^3 +bx^2 +a=0 may have a common root?

- anonymous

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- anonymous

when c = 0? hmm

- anonymous

I would appreciate if someone could solve this cause its been bugging me for days

- anonymous

or maybe if b = 0 and a & c are perfect cubes

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- anonymous

how did you get that?

- anonymous

get what?

- anonymous

your answers

- anonymous

you were asking for conditions so i was just doing trial and error

- anonymous

Could someone plz do this with full working thanks :D

- anonymous

@experimentX you seem like a genius, plz help

- experimentX

no I am not who said?

- anonymous

I have asked heaps of people but no one seems to know how to do it

- anonymous

well a 99 smart score speaks for itself ;)

- experimentX

just compare the coefficient
(ax^3+bx+c)/a=0, and (cx^3 +bx^2 +a)/c=0

- anonymous

ok what do i do when comparing the coefficients for the monic polynomial?

- anonymous

is it just c^2=a^2

- experimentX

the condition that they have same roots is that they must be same function or multiple of same function.

- anonymous

so is c^2=a^2 a condition?

- experimentX

woops!! looks like i misunderstood question ... how many common roots?

- anonymous

just says a common root, so 1 i guess?

- experimentX

let the common root be p ... q,r,s,t be other roots
f(x) = a(x-p)(x-q)(x-r)
g(x) = c(x-p)(x-s)(x-t)
equating to zero, you get
ax^3 - a(p+q+r)x^2 + (pq+qr+rp)ax + pqra = ax^3+bx+c
do the same for other, eliminate q, and r comparing coefficients
and from other eliminate s and t,
from both equations you get interms of p, equate 'p' from both and you have your condition. have to see if this get's complicated or not.

- hartnn

let that common root be 'y'
ay^3+by+c=0, and cy^3 +by^2 +a=0
eliminating 'y' from these two ! O.o

- experimentX

yeah that should work too ... solving cubic is real trouble though

- anonymous

what do you mean by eliminating e.g. like hartnn said eliminating 'y' from these two

- experimentX

the condition depends on your initial parameters like 'a', 'b', 'c'.
so you have to reduce it to the form in a,b,c which it must satisfy in order to have a common root.
you may use http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method to solve for cubic. just find one root and keep squaring or cubic unitl you get rid of surds.

- experimentX

if hartnn method didn't work then you can try mine, atmost you have to solve (two) quadratic equation ... i think.

- anonymous

how do i reduce it to the form in a,b,c?

- anonymous

soz, i get what you are talking about but i am still confused about how i should start :(

- experimentX

that's how i old you
ax^3 - a(p+q+r)x^2 + (pq+qr+rp)ax + pqra = ax^3+bx+c
p+q+r = 0
pq +qr +rp = b/a
pqra = c
------------------
from there equations eliminate q, and r ... you will get an equation in terms of p,a,b,c

- experimentX

similarly from here
c(x-p)(x-s)(x-t) = cx^3 +bx^2 +a
eliminate s and t, you should get equation in p,a,b,d
isolate p from both sides .. .this one and another one ... and equal them. you should get the required condition on 'a','b','c'

- experimentX

woops!! there has been slight error
ax^3 - a(p+q+r)x^2 + (pq+qr+rp)ax - pqra = ax^3+bx+c
there might be more erros, just take care of it.

- anonymous

how does this affect the three equations? and also when i eliminated the q and r i got a cubic of p

- experimentX

damn!! this looks like more hard way now.

- hartnn

mine is equally harder :\

- experimentX

let me try this on copy

- anonymous

let the common root be y then,
ay^3+by+c=0 ....i
cy^3+by^2+a=0 ...ii
subtract ii from i,
(a-c)y^3 + b(y-y^2) + (c-a)=0
(a-c)(y^3 -1) + by(1-y)=0
(a-c)(y-1)(y^2+y+1)-by(y-1)=0
(a-c) (y^2+y+1) -by=0
(a-c)y^2 + (a-c-b)y +(a-c)=0
Now, y exists only if (a-c-b)^2 >= 4*(a-c)*(a-c)

- anonymous

how did you get from (a-c) (y^2+y+1) -by=0 to (a-c)y^2 + (a-c-b)y +(a-c)=0?

- helder_edwin

simple distribution and factorization

- anonymous

yes

- helder_edwin

but why did u omit (y-1)? u cannot just divide by it

- anonymous

to simplify the expression (y-1) was omitted

- helder_edwin

i didn't say it doesn't. but u r excluding y=1 as a possible root.

- anonymous

oh I missed that...... when y=1 from equation (i) we get a+b+c=0
So, a+b+c=0 also works

- anonymous

Either a+b+c=0
OR
(a-b-c)^2>=4(a-c)^2
(a-b-c)>=2(a-c)
a+b-c>=0

- helder_edwin

i agree with the first one

- helder_edwin

but
\[ (-3)^2\geq(-2)^2\Rightarrow -3\geq-2 \]
?

- anonymous

oh cant use that when there is inequality

- anonymous

So, either a+b+c=0 OR (a-b-c)^2 >= 4(a-c)^2

- helder_edwin

yes. and the best i could do with your inequality is this:
\[ b^2-2b(a-c)\geq3(a-c)^2 \]
which doesn't look any better than yours

- ganeshie8

im trying to understand the solution... in the q they dint say common root needs to be real right ?

- ganeshie8

how we come up with det >= 0 ?

- anonymous

Because I assumed y is real

- ganeshie8

ohk we good to assume that ?

- helder_edwin

cubic equations have at least on real root. if we assume that it is their real roots that r equal then we r ok

- anonymous

They cannot only have only one common root imaginery...... if one common root is imaginery then both imaginery roots must be common.
But not sure how this would work

- helder_edwin

i was gonna say that.

- ganeshie8

yea 3 cases :-
1 common root (real)
2 common roots (imaginary)
3 common roots (1 real 2 imaginary)

- helder_edwin

in this case y=1 is no longer a problem. so we would go straight into the inequality
\[ (a-b-c)^2<4(a-c)^2 \]

- ganeshie8

that condition is oly a necessary condition, its not a sufficient condition
for ex, (a, b, c) = (1, 3, 2) wont give any common roots

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