polynomials question. (1 Viewer)

[toadstooltown]

I know this is really stupid and I should get this but
P(x)=x³+3x+2 with roots alpha, beta, gamma

find eqt. with roots alpha + 1/alpha etc. (a+1/a)

x=a+1/a,
a=(x+(x²-4)^½)/2 - right? =/
anyways, I tried subbing that in and expanding but it ain't working. should I just try and find alpha, beta, gamma then find the new roots and expand?

any help would be appreciated.

 

[vafa]

i have one solution but i do not know if u can use other ways:

solution:
in your original equation:
alpha+beta+gamma=0
alpha*beta+alpha*gamma+beta*gamma=3
alpha*beta*gamma==-2

generally i can write a cubic equation in the form:
ax^3+bx^2+cx+d=0 and
alpha+beta+gamma=-b/a
alpha*beta+alpha*gamma+beta*gamma=c/a
alpha*beta*gamma=-d/a

hence: alpha+1/alpha+beta+1/beta+gamma+1/gamma=-b/a
hence b=3a/2

(alpha+1/alpha)(beta+1/beta)+(alpha+1/alpha)(gamma+1/gamma)+(beta+1/beta)=c/a
but 0=c/a, hence c=0

(alpha+1/alpha)(beta+1/beta)(gamma+1/gamma)=-d/a
-4=-d/a, hence d=4a
if u loook at the 3 previous statements, b,d are in terms of a and evidently c=0 so:
ax^3 +(3a/2)x^2+4a=0
x^3+3/2x^2+4=0
eventually your eqyation is:
2x^3+3x^2+8=0

could u plz notify me if u got another solution?
thanks

 

[Yip]

Let X=(a+1)/a
aX-a=1
a=1/(X-1)

Since we kno a is a root, we ccan sub it into x^3+3x+2=0

1/(X-1)^3+3/(X-1)+2=0
2(X-1)^3+3(X-1)^2+1=0
2(X^3-3X^2+3X-1)+3(X^2-2X+1)+1=0
2X^3-6X^2+6X-2+3X^2-6X+3+1=0
2X^3-3X^2+2=0

So ur equation is 2x^3-3x^2+2=0 i think

 

[vafa]

you should have x=(a^2+1)/a not x=(a+1)/a, that is why your answer is not right.
and your way does not work in this question. i have tried that many times but unfortunately i could not get the answer.

 

[toadstooltown]

Ah, thanks, I did it with the general eqt. for cubic and eventually got it. Thanks for the help!

 

[toadstooltown]

Oh, vafa, my teacher it did it my way (letting alpha =[ x±(x²-4)^(½) ] / 2 ) and after a bit of working it came out anyways, I just some algebraic errors. Thanks for your help though!