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anonymous
 3 years 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?
anonymous
 3 years 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?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I would appreciate if someone could solve this cause its been bugging me for days

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or maybe if b = 0 and a & c are perfect cubes

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how did you get that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you were asking for conditions so i was just doing trial and error

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Could someone plz do this with full working thanks :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@experimentX you seem like a genius, plz help

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1no I am not who said?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I have asked heaps of people but no one seems to know how to do it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well a 99 smart score speaks for itself ;)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok what do i do when comparing the coefficients for the monic polynomial?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1the condition that they have same roots is that they must be same function or multiple of same function.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so is c^2=a^2 a condition?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1woops!! looks like i misunderstood question ... how many common roots?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just says a common root, so 1 i guess?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1let 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.

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1let 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
 3 years ago
Best ResponseYou've already chosen the best response.1yeah that should work too ... solving cubic is real trouble though

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what do you mean by eliminating e.g. like hartnn said eliminating 'y' from these two

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how do i reduce it to the form in a,b,c?

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1that'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
 3 years ago
Best ResponseYou've already chosen the best response.1similarly 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'

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1woops!! 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
 3 years ago
Best ResponseYou've already chosen the best response.0how does this affect the three equations? and also when i eliminated the q and r i got a cubic of p

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1damn!! this looks like more hard way now.

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1mine is equally harder :\

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1let me try this on copy

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0let 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)

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

helder_edwin
 3 years ago
Best ResponseYou've already chosen the best response.1simple distribution and factorization

helder_edwin
 3 years ago
Best ResponseYou've already chosen the best response.1but why did u omit (y1)? u cannot just divide by it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0to simplify the expression (y1) was omitted

helder_edwin
 3 years ago
Best ResponseYou've already chosen the best response.1i didn't say it doesn't. but u r excluding y=1 as a possible root.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Either a+b+c=0 OR (abc)^2>=4(ac)^2 (abc)>=2(ac) a+bc>=0

helder_edwin
 3 years ago
Best ResponseYou've already chosen the best response.1i agree with the first one

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh cant use that when there is inequality

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

helder_edwin
 3 years ago
Best ResponseYou've already chosen the best response.1yes. 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

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.0im trying to understand the solution... in the q they dint say common root needs to be real right ?

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.0how we come up with det >= 0 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Because I assumed y is real

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.0ohk we good to assume that ?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0They 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
 3 years ago
Best ResponseYou've already chosen the best response.1i was gonna say that.

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.0yea 3 cases : 1 common root (real) 2 common roots (imaginary) 3 common roots (1 real 2 imaginary)

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

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.0that 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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