Further maths core pure roots of polynomials question.

It's a 2018 paper that I can't seem to find a single markscheme for so I'd really appreciate if someone could help!

Every way I've tried either has too many unknowns or is so ridiculously long winded that I'm sure it's not right, so I'm more looking for a method that's time efficient. Is there something else you can infer from the information aside from the complex conjugate pair and equating a formula from each function?

I've attached a photo of it but in case that doesn't upload here's the written question:

f(x) = kx^2 + 3x - 11
g(x) = mx^3 - 2x^2 + 3x - 9

k and m are real constants.
The sum of the roots of f(x) is equal to the product of the roots of g(x).
g(x) has at least one root on imaginary axis.

Solve completely:

a) f(x) =0

b) g(x) = 0

Reply 1

Muttley79

20

Original post by KookieCrumble

It's a 2018 paper that I can't seem to find a single markscheme for so I'd really appreciate if someone could help!

Every way I've tried either has too many unknowns or is so ridiculously long winded that I'm sure it's not right, so I'm more looking for a method that's time efficient. Is there something else you can infer from the information aside from the complex conjugate pair and equating a formula from each function?

I've attached a photo of it but in case that doesn't upload here's the written question:

f(x) = kx^2 + 3x - 11
g(x) = mx^3 - 2x^2 + 3x - 9

k and m are real constants.
The sum of the roots of f(x) is equal to the product of the roots of g(x).
g(x) has at least one root on imaginary axis.

Solve completely:

a) f(x) =0

b) g(x) = 0

Can you post what you've tried? What paper is it?

Reply 2

DFranklin

18

Original post by KookieCrumble

Every way I've tried either has too many unknowns or is so ridiculously long winded that I'm sure it's not right, so I'm more looking for a method that's time efficient.

It's not hard to *verify* a solution to decide if it's correct, surely?

Is there something else you can infer from the information aside from the complex conjugate pair and equating a formula from each function?I'm not sure it you're using it and forgot to mention it, but knowing a complex root lies on the imaginary axis is going to simplify things quite a lot.

It's a little beyond something I can solve mentally, but it looks like using the standard equations relating the roots of a cubic to its coefficients give you a set of simultaneous equations that only take a few lines to solve.

Reply 3

KookieCrumble

OP

8

Original post by Muttley79

Can you post what you've tried? What paper is it?

It's S61294A further maths advanced paper 1 2018.

I've equated -3/k to 9/m
then subbed k into g(x) and formed:
the two equations of f(x) - defining to roots as alpha and beta
the three equations of g(x) - defining the roots as, a+bi, a-bi, c

but that still leaves me with the unknowns: alpha, beta, a, b, c, k, 6 unknowns to 5 equations so I'm one equation short of being able to solve it.

Are you supposed to infer that the roots of g(x) is some multiple of the roots of f(x)?
I then equated f(x) to g(x) but that doesn't seem to help much, you can't compare terms, really and when you get it onto one side its:

3kx^3 + (k+2)x^2 - 3 = 0

which doesn't seem to help??

edit: Ah, I might sub in the roots and then equate g to f and compare that way?
edit2: k so did ^ and seems to be working but still this is reeeally long

(edited 12 months ago)

Reply 4

ghostwalker

17

Original post by Muttley79

What paper is it?

Looks to be Edexcel Sample Assessment Material. Copyright is 2018. Paper ref: S61294A, and I suspect never published, only in draft form.

Page 19 in https://uploads-ssl.webflow.com/5f4d0d549c879fcac8e96b1f/5fe2b06713c6babbb3b39e3e_9FM0-01%20A%20level%20Core%20Pure%20Mathematics%201.pdf

Reply 5

ghostwalker

17

Original post by KookieCrumble

the three equations of g(x) - defining the roots as, a+bi, a-bi, c

Complex roots of g are on the imaginary axis, so what's "a? And carry on.

Reply 6

KookieCrumble

OP

8

Original post by ghostwalker

Complex roots of g are on the imaginary axis, so what's "a? And carry on.

but it doesn't specify that it's on the imaginary axis only, surely the root could be on the imaginary axis and the real axis. Am I confusing imaginary plane with axis then?

Reply 7

Muttley79

20

Original post by ghostwalker

Looks to be Edexcel Sample Assessment Material. Copyright is 2018. Paper ref: S61294A, and I suspect never published, only in draft form.

Page 19 in https://uploads-ssl.webflow.com/5f4d0d549c879fcac8e96b1f/5fe2b06713c6babbb3b39e3e_9FM0-01%20A%20level%20Core%20Pure%20Mathematics%201.pdf

Yes - it's not the same as the SAM on the Emporium.

Reply 8

ghostwalker

17

Original post by KookieCrumble

but it doesn't specify that it's on the imaginary axis only, surely the root could be on the imaginary axis and the real axis. Am I confusing imaginary plane with axis then?

Saying it's on the imaginary axis means it's on that specific line.

The plane is usually refered to as the complex plane, and the two axes are the real axis and the imaginary axis.

Reply 9

KookieCrumble

OP

8

Original post by ghostwalker

Saying it's on the imaginary axis means it's on that specific line.

The plane is usually refered to as the complex plane, and the two axes are the real axis and the imaginary axis.

ah thank you,

followed through on this, and have the roots to g but I think I've made a computation error as now the quadratic roots are imaginary lol. I'll redo that quick. either way thanks this makes it much easier!

Reply 10

Arconik

15

Done it. For those who score TSR looking for further maths questions to revise with like me here is the answers I got that seem to work with the initial conditions:

Spoiler

Reply 11

maddyali

1

Original post by Arconik

Done it. For those who score TSR looking for further maths questions to revise with like me here is the answers I got that seem to work with the initial conditions:

Spoiler

can you send the solutions for the question, it makes 0 sense what to do after m=-3k