anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
when c = 0? hmm
anonymous
  • anonymous
I would appreciate if someone could solve this cause its been bugging me for days
anonymous
  • anonymous
or maybe if b = 0 and a & c are perfect cubes

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anonymous
  • anonymous
how did you get that?
anonymous
  • anonymous
get what?
anonymous
  • anonymous
your answers
anonymous
  • anonymous
you were asking for conditions so i was just doing trial and error
anonymous
  • anonymous
Could someone plz do this with full working thanks :D
anonymous
  • anonymous
@experimentX you seem like a genius, plz help
experimentX
  • experimentX
no I am not who said?
anonymous
  • anonymous
I have asked heaps of people but no one seems to know how to do it
anonymous
  • anonymous
well a 99 smart score speaks for itself ;)
experimentX
  • experimentX
just compare the coefficient (ax^3+bx+c)/a=0, and (cx^3 +bx^2 +a)/c=0
anonymous
  • anonymous
ok what do i do when comparing the coefficients for the monic polynomial?
anonymous
  • anonymous
is it just c^2=a^2
experimentX
  • experimentX
the condition that they have same roots is that they must be same function or multiple of same function.
anonymous
  • anonymous
so is c^2=a^2 a condition?
experimentX
  • experimentX
woops!! looks like i misunderstood question ... how many common roots?
anonymous
  • anonymous
just says a common root, so 1 i guess?
experimentX
  • 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
  • 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
  • experimentX
yeah that should work too ... solving cubic is real trouble though
anonymous
  • anonymous
what do you mean by eliminating e.g. like hartnn said eliminating 'y' from these two
experimentX
  • 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
  • experimentX
if hartnn method didn't work then you can try mine, atmost you have to solve (two) quadratic equation ... i think.
anonymous
  • anonymous
how do i reduce it to the form in a,b,c?
anonymous
  • anonymous
soz, i get what you are talking about but i am still confused about how i should start :(
experimentX
  • 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
  • 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
  • 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
  • anonymous
how does this affect the three equations? and also when i eliminated the q and r i got a cubic of p
experimentX
  • experimentX
damn!! this looks like more hard way now.
hartnn
  • hartnn
mine is equally harder :\
experimentX
  • experimentX
let me try this on copy
anonymous
  • 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
  • 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
  • helder_edwin
simple distribution and factorization
anonymous
  • anonymous
yes
helder_edwin
  • helder_edwin
but why did u omit (y-1)? u cannot just divide by it
anonymous
  • anonymous
to simplify the expression (y-1) was omitted
helder_edwin
  • helder_edwin
i didn't say it doesn't. but u r excluding y=1 as a possible root.
anonymous
  • 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
  • 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
  • helder_edwin
i agree with the first one
helder_edwin
  • helder_edwin
but \[ (-3)^2\geq(-2)^2\Rightarrow -3\geq-2 \] ?
anonymous
  • anonymous
oh cant use that when there is inequality
anonymous
  • anonymous
So, either a+b+c=0 OR (a-b-c)^2 >= 4(a-c)^2
helder_edwin
  • 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
  • ganeshie8
im trying to understand the solution... in the q they dint say common root needs to be real right ?
ganeshie8
  • ganeshie8
how we come up with det >= 0 ?
anonymous
  • anonymous
Because I assumed y is real
ganeshie8
  • ganeshie8
ohk we good to assume that ?
helder_edwin
  • 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
  • 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
  • helder_edwin
i was gonna say that.
ganeshie8
  • ganeshie8
yea 3 cases :- 1 common root (real) 2 common roots (imaginary) 3 common roots (1 real 2 imaginary)
helder_edwin
  • 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
  • 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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