Polynomials question (or rather moving stuff about) (1 Viewer)

DVDVDVDV · Sep 25, 2011

x^3 + 3x + 2 = 0 has roots a, b and c. Find the equation with roots a + 1/a, b + 1/b and c + 1/c.

I've substituted ((x +/- sqrt(x^2 - 4))/2 in place of x, but the solutions for the Cambridge book skips several steps and goes straight to

+/-sqrt(x^2 - 4)(4x^2 + 8) = -4x^3 - 16.

I just can't work out how they got the above equation.

Help please?

K4M1N3 · Sep 26, 2011

Just sub in : x + 1/x and solve the polynomial....

EDIT: nvm that doesnt work....ive resorted to finding a + 1/a + b + 1/b + c + 1/c ...etc etc

btw if your doing the further questions: 6....the answer is 2x^3 + 3x^2 + 8 = 0 (from the book)

largarithmic · Sep 26, 2011

I assume they just subbed back in (x +/- root(x^2 - 4))/2 into the original equation and manipulated it using index laws/general algebra. You gotta be careful of the plusminus sign though but it should be alright? no tricks involved

the other way to do it would be to start from:
a+b+c = -3, ab+bc+ac = 0, abc = -2 and use Viete's formulae (Im not sure what they're called in school? 'product and sum of roots' or something?)

Youd get (a+1/a) + (b+1/b) + (c+1/c) = (a+b+c) + (1/a + 1/b + 1/c) = -3 + (ab+bc+ac)/abc = -3, so the x^2 term is 3.

(a+1/a)(b+1/b) + (a+1/a)(c+1/c) + (b+1/b)(c+1/c) = (ab+bc+ac) + (a/b + b/a + b/c + c/b + a/c + c/a) + (1/ab + 1/bc + 1/ac),
but (ab+bc+ac) = 0, and (a/b + b/a + b/c + c/b + a/c + c/a) = (a+b+c)(1/a + 1/b + 1/c) - 3 = (-3)(ab+ac+bc)/abc - 3 = (-3)(zero) - 3 = -3 and (1/ab + 1/bc + 1/ac) = (a+b+c)/abc = 3/2
So the x term is -3 + 3/2 = -3/2

And (a+1/a)(b+1/b)(c+1/c) = abc + 1/abc + ab/c + bc/a + ac/b + a/bc + b/ac + c/ab.
abc = -2, 1/abc = -1/2. Then ab/c + bc/a + ac/b = (ab+bc+ac)(1/a+1/b+1/c) - 2(a+b+c) = (-3)(zero, we've found 1/a +1/b+1/c before) - 2(-3) = 6. And also a/bc + b/ac + c/ab = (a+b+c)(1/ab + 1/bc + 1/ac) - (a/b + b/a + b/c + c/b + a/c + c/a) = (-3)(3/2, found this before) - (-3, found this before) = 3 - 9/2 = -3/2.

So the constant term is -(-2 -1/2 + 6 - 3/2) =- 2.

This gives an answer of x^3 + 3x - 3/2 x - 2 = 0, I have no idea whether thats right though coz this method could easily mess up the arithmetic. But this method will almost always work with enough algebraic manipulation; from the three original polynomial root/coefficient equations you can derive almost any symmetric expression in a,b,c; including the root/coefficient equations for the polynomial you want to form.

DVDVDVDV · Sep 26, 2011

Yeah, they subbed it in, but I can't do it with the plusminus sign unfortunately

largarithmic · Sep 26, 2011

DVDVDVDV said:
Yeah, they subbed it in, but I can't do it with the plusminus sign unfortunately

its not that bad, just act as if its a normal plus or minus sign. When you multiply a plusminus sign by a plusminus sign (i.e. you square it) you get a plus, and I think thats all you really need to know. Or just split it into two cases, 1. when its a plus, 2. when its a minus etc

math man · Oct 5, 2011

so you should work out that you have to sub a = (1/2)[x + sqrt(x^2 - 4)] into the original eqn...
doing so gives:
(1/8)[x + sqrt(x^2 - 4)]^3 + (3/2)[x + sqrt(x^2 - 4)] + 2 = 0 ...now i will times everything by 8
[x + sqrt(x^2 - 4)]^3 + 12[x + sqrt(x^2 - 4)] + 16 = 0 ...here expanding this cubic is nasty so what you should do is factorise the [x + sqrt(x^2 - 4)] from the first two terms
[x + sqrt(x^2 - 4)]{[x + sqrt(x^2 - 4)]^2 + 12]} + 16 = 0 ...now you expand the squared
[x + sqrt(x^2 - 4)]{x^2 + 2xsqrt(x^2 - 4) + x^2 - 4 + 12} + 16 = 0 ...now simply this
[x + sqrt(x^2 - 4)]{2x^2 + 2xsqrt(x^2 -4) +8} + 16 = 0 ...now im afraid we expand this..but this is easier than expanding the cubic
2x^3 + 2x^2sqrt(x^2 -4) + 8x + 2x^2sqrt(x^2 -4) + 2x(x^2 -4) + 8sqrt(x^2 -4) + 16 = 0 ...simply this obtains
2x^3 + 4x^2sqrt(x^2 -4) + 8x + 2x^3 -8x + 8sqrt(x^2 -4) + 16 = 0 ....simply this again
4x^3 + 4x^2sqrt(x^2 -4) + 8sqrt(x^2 -4) + 16 = 0 .....divide everything by 4 and factorise the sqrt(x^2 -4) from the 2 terms...
x^3 + 4 + sqrt(x^2 -4)[x^2 + 2] = 0 ...take the sqrt(x^2 -4) part to the other side and square both sides to finally eliminate the sqrt
(x^3 + 4)^2 = [-sqrt(x^2 -4)(x^2 + 2)]^2 ....expand this gives
x^6 + 8x^3 + 16 = (x^2 - 4)(x^4 + 4x^2 + 4) ...dont be scared with these x^6 and x^4 they will cancel as follows...
x^6 + 8x^3 + 16 = x^6 + 4x^4 +4x^2 - 4x^4 - 16x^2 - 16 ....simplyfying this gives...
8x^3 +12x^2 + 32 = 0 ...dividing through by 4 gives
2x^3 + 3x^2 + 8 = 0, which is the answer....this was very long working so dont be afraid if it takes long..this was a messy question

Alkanes · Oct 5, 2011

lol you guys should learn how to use latex

math man · Oct 5, 2011

so you should work out that you have to sub $a = (1/2)[x + \sqrt{x^2 - 4}]$
into the original eqn...
doing so gives:
$(1/8)[x + \sqrt{x^2 - 4}]^3 + (3/2)[x + \sqrt{x^2 - 4}] + 2 = 0$
...now i will times everything by 8
$[x + \sqrt{x^2 - 4}]^3 + 12[x + \sqrt{x^2 - 4}] + 16 = 0$
...here expanding this cubic is nasty so what you should do is factorise the $[x + \sqrt{x^2 - 4}]$
from the first two terms
$[x + \sqrt{x^2 - 4}]{[x + \sqrt{x^2 - 4}]^2 + 12]} + 16 = 0$
...now you expand the squared part
$[x + \sqrt{x^2 - 4]}[{x^2 + 2x\sqrt{x^2 - 4} + x^2 - 4 + 12}] + 16 = 0$
...now simply this
$[x + \sqrt{x^2 - 4}][{2x^2 + 2x\sqrt{x^2 -4} +8}] + 16 = 0$
...now im afraid we expand this..but this is easier than expanding the cubic
$2x^3 + 2x^2\sqrt{x^2 -4 }+ 8x + 2x^2\sqrt{x^2 -4} + 2x(x^2 -4) + 8\sqrt{x^2 -4} + 16 = 0$
...simply this obtains
$2x^3 + 4x^2\sqrt{x^2 -4} + 8x + 2x^3 -8x + 8\sqrt{x^2 -4} + 16 = 0$
....simply this again
$4x^3 + 4x^2\sqrt{x^2 -4} + 8\sqrt{x^2 -4} + 16 = 0$
.....divide everything by 4 and factorise the $\sqrt{x^2 -4}$
from the 2 terms...
$x^3 + 4 + \sqrt{x^2 -4}[x^2 + 2] = 0$
...take the sqrt(x^2 -4) part to the other side and square both sides to finally eliminate the sqrt part
$(x^3 + 4)^2 = [-\sqrt{x^2 -4}(x^2 + 2)]^2$
....expand this gives
$x^6 + 8x^3 + 16 = (x^2-4)(x^4 + 4x^4 + 4)$

...dont be scared with these x^6 and x^4 they will cancel as follows...
$x^6 + 8x^3 + 16 = x^6 + 4x^4 +4x^2 - 4x^4 - 16x^2 - 16$
....simplyfying this gives...
$8x^3 +12x^2 + 32 = 0$
...dividing through by 4 gives
$2x^3 + 3x^2 + 8 = 0$
, which is the answer....this was very long working so dont be afraid if it takes long..this was a messy question

math man · Oct 5, 2011

is that better

seanieg89 · Oct 6, 2011

math man · Oct 6, 2011

seanieg89 said:
$Although it leads you to the correct answer for this particular polynomial, simply substituting $ \frac{x+\sqrt{x^2-4}}{2}$ into your polynomial is a little dodgy since: $\\y=x+x^{-1}$ does \emph{not} necessarily imply that $ x=\frac{y+\sqrt{y^2-4}}{2}.\\$ So you run into problems for polynomials that have a root at say $x=1/2$. To make the argument rigorous you would have to do something like construct a polynomial by multiplying $ P\left(\frac{x+\sqrt{x^2-4}}{2}\right)$ by $P\left(\frac{x-\sqrt{x^2-4}}{2}\right) $ or by making careful use of the '$\pm$' sign. It is probably best to just use Vieta's formulae here.$

subbing either the + or - root doesn't matter..they both give right answer..that's why i didnt bother with the neg one

seanieg89 · Oct 6, 2011

My point is that you only end up with an equation that is satisfied at (a+1/a, b+1/b, c+1/c) after the step where you square both sides to eliminate the square root sign.
It is not obvious that this method is valid without use of the +- sign.

If you try the same method for say:
P(x)=x^3+3x-13/8 (which has root 1/2)

Then you will find that your equations before the squaring step i mentioned are not satisfied at x=2+1/2=5/2.

math man · Oct 6, 2011

seanieg89 said:
My point is that you only end up with an equation that is satisfied at (a+1/a, b+1/b, c+1/c) after the step where you square both sides to eliminate the square root sign.
It is not obvious that this method is valid without use of the +- sign.

If you try the same method for say:
P(x)=x^3+3x-13/8 (which has root 1/2)

Then you will find that your equations before the squaring step i mentioned are not satisfied at x=2+1/2=5/2.

it doesnt matter which you use, because we have $a = (x \pm \sqrt{x^2 - 4})/2$ , where a is our root of the new eqn,
as a is our root P(a)= 0, so that means P( $a = (x + \sqrt{x^2 - 4})/2$ )=P( $a = (x - \sqrt{x^2 - 4})/2$ )=0
so you can use either cause both will give right answer

seanieg89 · Oct 6, 2011

math man · Oct 6, 2011

i know where you are coming from lol...im just trying to point out both the + or - solution will give the same answer as they are both roots of the polynomial so you only need to sub in one of them to find the polynomial with the new desired roots

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