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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?
 9 months ago
 9 months ago
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?
 9 months ago
 9 months ago

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aceaceBest ResponseYou've already chosen the best response.0
I would appreciate if someone could solve this cause its been bugging me for days
 9 months ago

xlegendxBest ResponseYou've already chosen the best response.0
or maybe if b = 0 and a & c are perfect cubes
 9 months ago

xlegendxBest ResponseYou've already chosen the best response.0
you were asking for conditions so i was just doing trial and error
 9 months ago

aceaceBest ResponseYou've already chosen the best response.0
Could someone plz do this with full working thanks :D
 9 months ago

aceaceBest ResponseYou've already chosen the best response.0
@experimentX you seem like a genius, plz help
 9 months ago

experimentXBest ResponseYou've already chosen the best response.1
no I am not who said?
 9 months ago

aceaceBest ResponseYou've already chosen the best response.0
I have asked heaps of people but no one seems to know how to do it
 9 months ago

aceaceBest ResponseYou've already chosen the best response.0
well a 99 smart score speaks for itself ;)
 9 months ago

experimentXBest ResponseYou've already chosen the best response.1
just compare the coefficient (ax^3+bx+c)/a=0, and (cx^3 +bx^2 +a)/c=0
 9 months ago

aceaceBest ResponseYou've already chosen the best response.0
ok what do i do when comparing the coefficients for the monic polynomial?
 9 months ago

experimentXBest ResponseYou've already chosen the best response.1
the condition that they have same roots is that they must be same function or multiple of same function.
 9 months ago

aceaceBest ResponseYou've already chosen the best response.0
so is c^2=a^2 a condition?
 9 months ago

experimentXBest ResponseYou've already chosen the best response.1
woops!! looks like i misunderstood question ... how many common roots?
 9 months ago

aceaceBest ResponseYou've already chosen the best response.0
just says a common root, so 1 i guess?
 9 months ago

experimentXBest ResponseYou've already chosen the best response.1
let the common root be p ... q,r,s,t be other roots f(x) = a(xp)(xq)(xr) g(x) = c(xp)(xs)(xt) 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.
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
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
 9 months ago

experimentXBest ResponseYou've already chosen the best response.1
yeah that should work too ... solving cubic is real trouble though
 9 months ago

aceaceBest ResponseYou've already chosen the best response.0
what do you mean by eliminating e.g. like hartnn said eliminating 'y' from these two
 9 months ago

experimentXBest ResponseYou've already chosen the best response.1
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.
 9 months ago

experimentXBest ResponseYou've already chosen the best response.1
if hartnn method didn't work then you can try mine, atmost you have to solve (two) quadratic equation ... i think.
 9 months ago

aceaceBest ResponseYou've already chosen the best response.0
how do i reduce it to the form in a,b,c?
 9 months ago

aceaceBest ResponseYou've already chosen the best response.0
soz, i get what you are talking about but i am still confused about how i should start :(
 9 months ago

experimentXBest ResponseYou've already chosen the best response.1
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
 9 months ago

experimentXBest ResponseYou've already chosen the best response.1
similarly from here c(xp)(xs)(xt) = 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'
 9 months ago

experimentXBest ResponseYou've already chosen the best response.1
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.
 9 months ago

aceaceBest ResponseYou've already chosen the best response.0
how does this affect the three equations? and also when i eliminated the q and r i got a cubic of p
 9 months ago

experimentXBest ResponseYou've already chosen the best response.1
damn!! this looks like more hard way now.
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
mine is equally harder :\
 9 months ago

experimentXBest ResponseYou've already chosen the best response.1
let me try this on copy
 9 months ago

sauravshakyaBest ResponseYou've already chosen the best response.4
let the common root be y then, ay^3+by+c=0 ....i cy^3+by^2+a=0 ...ii subtract ii from i, (ac)y^3 + b(yy^2) + (ca)=0 (ac)(y^3 1) + by(1y)=0 (ac)(y1)(y^2+y+1)by(y1)=0 (ac) (y^2+y+1) by=0 (ac)y^2 + (acb)y +(ac)=0 Now, y exists only if (acb)^2 >= 4*(ac)*(ac)
 9 months ago

aceaceBest ResponseYou've already chosen the best response.0
how did you get from (ac) (y^2+y+1) by=0 to (ac)y^2 + (acb)y +(ac)=0?
 9 months ago

helder_edwinBest ResponseYou've already chosen the best response.1
simple distribution and factorization
 9 months ago

helder_edwinBest ResponseYou've already chosen the best response.1
but why did u omit (y1)? u cannot just divide by it
 9 months ago

sauravshakyaBest ResponseYou've already chosen the best response.4
to simplify the expression (y1) was omitted
 9 months ago

helder_edwinBest ResponseYou've already chosen the best response.1
i didn't say it doesn't. but u r excluding y=1 as a possible root.
 9 months ago

sauravshakyaBest ResponseYou've already chosen the best response.4
oh I missed that...... when y=1 from equation (i) we get a+b+c=0 So, a+b+c=0 also works
 9 months ago

sauravshakyaBest ResponseYou've already chosen the best response.4
Either a+b+c=0 OR (abc)^2>=4(ac)^2 (abc)>=2(ac) a+bc>=0
 9 months ago

helder_edwinBest ResponseYou've already chosen the best response.1
i agree with the first one
 9 months ago

helder_edwinBest ResponseYou've already chosen the best response.1
but \[ (3)^2\geq(2)^2\Rightarrow 3\geq2 \] ?
 9 months ago

sauravshakyaBest ResponseYou've already chosen the best response.4
oh cant use that when there is inequality
 9 months ago

sauravshakyaBest ResponseYou've already chosen the best response.4
So, either a+b+c=0 OR (abc)^2 >= 4(ac)^2
 9 months ago

helder_edwinBest ResponseYou've already chosen the best response.1
yes. and the best i could do with your inequality is this: \[ b^22b(ac)\geq3(ac)^2 \] which doesn't look any better than yours
 9 months ago

ganeshie8Best ResponseYou've already chosen the best response.0
im trying to understand the solution... in the q they dint say common root needs to be real right ?
 9 months ago

ganeshie8Best ResponseYou've already chosen the best response.0
how we come up with det >= 0 ?
 9 months ago

sauravshakyaBest ResponseYou've already chosen the best response.4
Because I assumed y is real
 9 months ago

ganeshie8Best ResponseYou've already chosen the best response.0
ohk we good to assume that ?
 9 months ago

helder_edwinBest ResponseYou've already chosen the best response.1
cubic equations have at least on real root. if we assume that it is their real roots that r equal then we r ok
 9 months ago

sauravshakyaBest ResponseYou've already chosen the best response.4
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
 9 months ago

helder_edwinBest ResponseYou've already chosen the best response.1
i was gonna say that.
 9 months ago

ganeshie8Best ResponseYou've already chosen the best response.0
yea 3 cases : 1 common root (real) 2 common roots (imaginary) 3 common roots (1 real 2 imaginary)
 9 months ago

helder_edwinBest ResponseYou've already chosen the best response.1
in this case y=1 is no longer a problem. so we would go straight into the inequality \[ (abc)^2<4(ac)^2 \]
 9 months ago

ganeshie8Best ResponseYou've already chosen the best response.0
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
 9 months ago
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