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A level further maths problem - roots of quartic equation

I attempted this question and was able to get as far as:

α²r³ = 1/32

I can't see a way of solving this problem without guessing values of α and r, can anyone see another way? Many thanks.
ex4c q9.png

Reply 1

flaurie

Volunteer Team

Original post by TSRTD

This one's really stumped me - I keep going round in circles getting α²r³ = 1/32 like you
Bumping the thread so someone else might see it!

Reply 2

mqb2766

Original post by TSRTD

The cubic coefficient will give the sum of a geometric sequence (roots) as well. Just solve those two equations for r and alpha.

(edited 3 years ago)

Reply 3

TSRTD

Original post by mqb2766

The cubic coefficient will give the sum of a geometric sequence (roots) as well. Just solve those two equations for r and alpha.

I tried this, and I think I encountered the same problem as I have with other routes. Everything I try leads to having to solve a quartic or quintic equation, neither of which I have a rigorous method for solving.

Reply 4

DFranklin

As I said in the other thread, if you form those 2 equations, there's an obvious guess for r and alpha. It's obvious "enough" that I very much expect you were just supposed to spot it.

If you're worried about rigour, a quartic has at most 4 roots, so if you find 4 roots that work, you have solved it, even if you did it totally by trial and error.

Reply 5

Interea

Original post by TSRTD

Knowing Edexcel they probably just expect you to guess and then check it works. So in this case if we know α²r³ = 1/32 = (1/2)⁵, so it looks like having α=r=1/2 will work (and I think if you check it then it does). I don't see any way to do it other than making a logical guess from this point :confused:

Reply 6

old_engineer

Original post by TSRTD

The Pearson solution bank notes that 1/32 = 1/(2^5) and goes from there to offer r = 1/2 and a = 1/2 as candidate solutions which are then checked against the formulas for one or more of the polynomial coefficients. Any other approach looks quite difficult to me.

Edit: overlap with posts #5 and #6 acknowledged.

(edited 3 years ago)

Reply 7

MiladA

Original post by TSRTD

Four roots, x = α, rα, r^2*α, r^3*a. Could factor the quartic as (x - α)(x - rα)(x - r^2*α)(x - r^3*α} then expand and compare coefficients.

Or you could use sum of the roots/products of roots formulae?

Reply 8

DFranklin

Original post by old_engineer

Not really any "better", but by the rational root theorem, any rational root p/q (with p,q in lowest terms) has p | 3 and q | 3072. You can rule out 3 dividing p or q easily and you're then down to a very small number of possibilities.

[Of course, you don't know the roots are rational, which is why this isn't perhaps as good as it looks].

Reply 9

mqb2766

After too much time, the best I can get is using (mainly) the quadratic coefficient you can get
30 = (31 - 4r^2)sqrt(2r)
Which, by inspection, has the two solutions r=2,1/2. It takes too long to get it in that form however.

Reply 10

mqb2766

Last effort which is probably closer to a proper solution.

Let the centred geometric sequence have terms
m/s^3, m/s, ms, ms^3
Then the geometric mean, m, is 1/4sqrt(2). That's as before, just taking the fourth root of the constant term. Note r = s^2 now.

Looking at the quadratic coefficient we get
m^2(s^-2 + s^2)^2 + m^2(s^-2 + s^2) = 35/128
We can solve that quadratic getting
(s^-2 + s^2) = 5/2
And solving that quadratic gives
s^2 = 2, 1/2

So sorted. r=1/2, alpha=1/2 or alpha=1/16, r=2.

Reply 11

TSRTD

Original post by mqb2766

Thank you for this! I like this solution.
The quartic equation that I had in terms of r after messing around with the coefficient equations gave two real solutions of r = 1/2 and r = 2, both of which make sense as having r = 2 and r = 1/2 results in identical roots.

Reply 12

mqb2766

Original post by TSRTD

There are two solutions for the same geometric sequence 1/16,1/8,1/4,1/2. One increases (r=2, alpha=1/16), the other decreases (r=1/2,alpha=1/2). They are both solutions to this quartic, just starting/indexed differently.

Note, you don't have to use a centred geometric sequence, but it's easier to spot the factorisation for the quadratic part. I kept pulling the wrong power of r out when doing it in the original geometric form. Similarly, the centred form has the nice property that m is the geometric mean and the (4th) root of the constant.