Challenge Polynomials problem for current students (1 Viewer)

[Xayma]

Too bad the time I spent on that wouldn't have been worth it in a test

 

[Grey Council]

heh, i'll take Buchanan's role:
You learn more from the questions you DON'T get out, then from the questions you do.
cause the questions you can't get out, you find out what doesn't work.

 

[CM_Tutor]

Originally posted by Xayma
And thats what Im trying to show. But I cant find any Wait if b=c=d=0 ax^4=0 only has one root, So it isnt true to say...

The problem with that answer is that the proof originally discounted x = 0 as a root, and thus d <> 0. However, the question can still be easily done.

Take a set of values, like a = d = 1, and b = c = 2, which satisfy the required result bc = 4ad. We know that, if it has a triple root, then it's at x = -b / 2a = -2 / 2(1) = -1. But x = -1 does not satisfy x⁴ + 2x³ + 2x + 1 = 0. Therefore there is no triple root, and so the answer is "no".

To check, you could solve x⁴ + 2x³ + 2x + 1 = 0 (difficult) or sketch y = x⁴ + 2x³ + 2x + 1, which has only two roots, neither of them a triple root (as y <> 0 when y'' = 0).

If anyone is interested, x⁴ + 2x³ + 2x + 1 factorises to [x² + (1 + sqrt(3))x + 1][x² + (1 - sqrt(3))x + 1], and the solutions of x⁴ + 2x³ + 2x + 1 = 0 are:
x = [-1 - sqrt(3) +/- 4throot(12)] / 2 and x = [sqrt(3) - 1 +/- i * 4throot(12)] / 2

 

[Xayma]

Yeah but I left the proof, after you gave me that hint but I like your way better.

 

[Grey Council]

for convenience:
1. If P(x) = Ax⁴ + Bx³ + Cx² + Dx + E is an even polynomial, and one of the roots of P(x) = 0 is twice another of the roots, prove that 4C² = 25AE

2. The equation x³ + 3x - 5 = 0 has roots at alpha, beta and gamma (which, for clarity, I'm going to write as a, b, and c.). Find an equation with roots ab / c, bc / a and ca / b.

 

[grimreaper]

ok I just tried again for the second of those newer questions and got x^3 - 9x^2 - 150x - 625 which seems completely wrong lol. I did it by saying that the roots can be written as abc/c^2, abc/b^2 and abc/c^2 and you can replace abc with 5... is that the right way to do it?

 

[CM_Tutor]

Originally posted by grimreaper
I did it by saying that the roots can be written as abc/c^2, abc/b^2 and abc/c^2 and you can replace abc with 5... is that the right way to do it?

Yes, that's the 4u 'short cut', but there's still an algebra problem to get around from there. The correct answer is posted elsewhere in this thread, so you'll know when you get it right.

 

[DcM]

so 5/c^2, 5/b^2, 5/a^2 are the roots of the new eqn rite..
so u just let y = 5/x^2
and sub bak into the original eqn
expand and simplify and u ll get the answer =)

 

[CM_Tutor]

We still don't have a solution for:

1. If P(x) = Ax⁴ + Bx³ + Cx² + Dx + E is an even polynomial, and one of the roots of P(x) = 0 is twice another of the roots, prove that 4C² = 25AE

 

[Grey Council]

hrm, one thing i've always meant to ask someone is this:
if that particular polynomial is EVEN, then does that mean that B=0 and D=0?

 

[CM_Tutor]

Originally posted by Grey Council
hrm, one thing i've always meant to ask someone is this:
if that particular polynomial is EVEN, then does that mean that B=0 and D=0?

Yes, it does, and it's very easy to prove... put P(x) = P(-x), expand, cancel and equate coefficients.

And, this is one of the important realisations to solving this problem quickly - although if you don't realise that we know that B = D = 0, it should still come out, as you'll end up concluding this anyway.

 

[nike33]

1. If P(x) = Ax4 + Bx3 + Cx2 + Dx + E is an even polynomial, and one of the roots of P(x) = 0 is twice another of the roots, prove that 4C2 = 25AE

let roots be s, -s, 2s, -2s

using product rule, sum of product etc... for Ax4 + Cx2 + E

u get

s^2 = -c / 5A (1)

and

4s^4 = E / A (2)

by sub (1) into (2)..u get 4C^2 = 25AE


...i hope this is right...

 

[nike33]

is the correct answer but for a typo in the last line - it should be x3 - 9x2 / 5 - 6x - 5 = 0, or alternately,
5x3 - 9x2 - 30x - 25 = 0

Extn 2 people - can you find an easier way to do this?

sub in x = +-sqr(5/y)

as roots

ab / c, bc / a and ca / b. let the roots be abc/c^2, abc/a^2, abc/b^2. abc = 5 (product of roots)

hence qn can be rewitten as find roots of (5/a^2), (5/b^2), (5/c^2) ..now y = (5 / x^2) hence x = +-sqr(5/y)

sub this into the original eqn and u get the req ans..