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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?

Mathematics
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when c = 0? hmm
I would appreciate if someone could solve this cause its been bugging me for days
or maybe if b = 0 and a & c are perfect cubes

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Other answers:

how did you get that?
get what?
your answers
you were asking for conditions so i was just doing trial and error
Could someone plz do this with full working thanks :D
@experimentX you seem like a genius, plz help
no I am not who said?
I have asked heaps of people but no one seems to know how to do it
well a 99 smart score speaks for itself ;)
just compare the coefficient (ax^3+bx+c)/a=0, and (cx^3 +bx^2 +a)/c=0
ok what do i do when comparing the coefficients for the monic polynomial?
is it just c^2=a^2
the condition that they have same roots is that they must be same function or multiple of same function.
so is c^2=a^2 a condition?
woops!! looks like i misunderstood question ... how many common roots?
just says a common root, so 1 i guess?
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.
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
yeah that should work too ... solving cubic is real trouble though
what do you mean by eliminating e.g. like hartnn said eliminating 'y' from these two
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.
if hartnn method didn't work then you can try mine, atmost you have to solve (two) quadratic equation ... i think.
how do i reduce it to the form in a,b,c?
soz, i get what you are talking about but i am still confused about how i should start :(
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
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'
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.
how does this affect the three equations? and also when i eliminated the q and r i got a cubic of p
damn!! this looks like more hard way now.
mine is equally harder :\
let me try this on copy
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)
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?
simple distribution and factorization
yes
but why did u omit (y-1)? u cannot just divide by it
to simplify the expression (y-1) was omitted
i didn't say it doesn't. but u r excluding y=1 as a possible root.
oh I missed that...... when y=1 from equation (i) we get a+b+c=0 So, a+b+c=0 also works
Either a+b+c=0 OR (a-b-c)^2>=4(a-c)^2 (a-b-c)>=2(a-c) a+b-c>=0
i agree with the first one
but \[ (-3)^2\geq(-2)^2\Rightarrow -3\geq-2 \] ?
oh cant use that when there is inequality
So, either a+b+c=0 OR (a-b-c)^2 >= 4(a-c)^2
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
im trying to understand the solution... in the q they dint say common root needs to be real right ?
how we come up with det >= 0 ?
Because I assumed y is real
ohk we good to assume that ?
cubic equations have at least on real root. if we assume that it is their real roots that r equal then we r ok
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
i was gonna say that.
yea 3 cases :- 1 common root (real) 2 common roots (imaginary) 3 common roots (1 real 2 imaginary)
in this case y=1 is no longer a problem. so we would go straight into the inequality \[ (a-b-c)^2<4(a-c)^2 \]
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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