# Further Maths - Roots of Polynomials

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#1
Hello, I have this question:

I have used the product of all 3 roots equalling -d/a relationship to then further solve for alpha. I end up solving a quadratic to which alpha is 3 - 2i
I then go to substitute 3 2i into the 3 expressions of the roots of the cubic and I don’t get the correct answers because one of them should be the complex conjugate of 3 2i -> 3 - 2i.
I have checked and made sure the answers to alpha are solved correct. I don’t understand why substitution into the expressions for solved alpha doesn’t yield me the other two correct roots if given those are correct expressions for the 3 roots.
My workings:

Any help is appreciated, thanks
Last edited by BrandonS15; 6 months ago
1
6 months ago
#2
Have you tried substituting 3 2i into f(x) to check if it’s actually a root? If it isn’t, you know the mistake is in the algebra before that. If it is, you can assume 3-2i will also work and use those to find the other.
Edit: just realised this will be awkward with m and n, but I would still do it for the and - and compare answers.
Last edited by Dancer2001; 6 months ago
0
6 months ago
#3
(Original post by BrandonS15)
Hello, I have this question:

I have used the product of all 3 roots equalling -d/a relationship to then further solve for alpha. I end up solving a quadratic to which alpha is 3 +- 2i
I then go to substitute 3 + 2i into the 3 expressions of the roots of the cubic and I don’t get the correct answers because one of them should be the complex conjugate of 3 + 2i -> 3 - 2i.
I have checked and made sure the answers to alpha are solved correct. I don’t understand why substitution into the expressions for solved alpha doesn’t yield me the other two correct roots if given those are correct expressions for the 3 roots.
My workings:

Any help is appreciated, thanks
Images don't appear.
0
6 months ago
#4
(Original post by BrandonS15)
Hello, I have this question:

I have used the product of all 3 roots equalling -d/a relationship to then further solve for alpha. I end up solving a quadratic to which alpha is 3 +- 2i
I then go to substitute 3 + 2i into the 3 expressions of the roots of the cubic and I don’t get the correct answers because one of them should be the complex conjugate of 3 + 2i -> 3 - 2i.
I have checked and made sure the answers to alpha are solved correct. I don’t understand why substitution into the expressions for solved alpha doesn’t yield me the other two correct roots if given those are correct expressions for the 3 roots.
My workings:

Any help is appreciated, thanks
Roots of polynomials only come in conjugate pairs if all of the coefficents are real. I haven't done the second part of the question, but I expect you will find that at least one of m and n are not real.
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6 months ago
#5
Hey, I know it's not very helpful, but I just wanted to say:
"Well done for maintaining such a neat handwriting".
Keep it up :P
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#6
(Original post by RDKGames)
Images don't appear.
Question:

Workings:
0
6 months ago
#7
(Original post by BrandonS15)
Question:

Workings:
I guess TSR is having a slow day.

0
6 months ago
#8
(Original post by RDKGames)
I guess TSR is having a slow day.

I can see them all okay
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6 months ago
#9
(Original post by Sir Cumference)
I can see them all okay
Me too.
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#10
(Original post by Pangol)
Roots of polynomials only come in conjugate pairs if all of the coefficents are real. I haven't done the second part of the question, but I expect you will find that at least one of m and n are not real.
I checked the algebra and its correct in yielding alpha as 3 +- 2i however I understand that the complex roots only come as conjugate pairs if the coefficients of the polynomial are real. Therefore instead of assuming they should be conjugate pairs, I directly substituted alpha as 3 + 2i into 1/alpha and the other expression to yield the other two given roots to the cubic however, these do not match the correct answers of 3 + 2i, 3 -2i and 4. I’m not sure how to obtain the other two roots of the cubic given the expression of the roots and a correctly calculated value for alpha.
0
6 months ago
#11
(Original post by BrandonS15)
I checked the algebra and its correct in yielding alpha as 3 +- 2i however I understand that the complex roots only come as conjugate pairs if the coefficients of the polynomial are real. Therefore instead of assuming they should be conjugate pairs, I directly substituted alpha as 3 + 2i into 1/alpha and the other expression to yield the other two given roots to the cubic however, these do not match the correct answers of 3 + 2i, 3 -2i and 4. I’m not sure how to obtain the other two roots of the cubic given the expression of the roots and a correctly calculated value for alpha.
Are you sure those are the correct answers? Where does the question come from?
0
6 months ago
#12
(Original post by Pangol)
Are you sure those are the correct answers?
I think there may be typos in the question, and suspect (tentatively) that the roots are supposed to be a, 13/a and a + 13/a - 2.
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6 months ago
#13
(Original post by Pangol)
Are you sure those are the correct answers? Where does the question come from?
Oh hang on, I've found it. This is one of those Edexcel end of unit tests that they put out when the A Levels were reformed. There are quite a few errors in them, and I think this question is one of them.

It looks to me as if you have done the right thing based on the information given to you, but the way the question is phrased is wrong. I'm trying to see what their error is, but can't come up with anything that puts it quite right.
0
#14
(Original post by Pangol)
Oh hang on, I've found it. This is one of those Edexcel end of unit tests that they put out when the A Levels were reformed. There are quite a few errors in them, and I think this question is one of them.

It looks to me as if you have done the right thing based on the information given to you, but the way the question is phrased is wrong. I'm trying to see what their error is, but can't come up with anything that puts it quite right.
Yes it is Unit Test 5 for the Core Pure Book 1 FM
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#15
(Original post by old_engineer)
I think there may be typos in the question, and suspect (tentatively) that the roots are supposed to be a, 13/a and a + 13/a - 2.
Using those 3 expressions as the given expressions in the questions do yield me the correct roots and therefore the correct values of m and n as given in the answers. I’m now confident its an error in the question. Thanks
0
6 months ago
#16
(Original post by Sir Cumference)
I can see them all okay
It works now that I switched to the 'Beta' version of the new page layout (enabled all the way at the bottom of the page).

Perhaps this is a sign that I should be getting used to it
0
6 months ago
#17
(Original post by old_engineer)
I think there may be typos in the question, and suspect (tentatively) that the roots are supposed to be a, 13/a and a + 13/a - 2.
Oh you genius. I tried using 13/a instead of 1/a, but couldn't see how to adjust the other root to match. This must be it. (PRSOM)
0
6 months ago
#18
(Original post by RDKGames)
It works now that I switched to the 'Beta' version of the new page layout (enabled all the way at the bottom of the page).

Perhaps this is a sign that I should be getting used to it
I'm surprised that they haven't got rid of the old page yet - I'm sure they said they would last year!
1
6 months ago
#19
(Original post by Sir Cumference)
I'm surprised that they haven't got rid of the old page yet - I'm sure they said they would last year!
Oh you know how they are, they say they will do something but take 10 years to touch the subject
0
5 months ago
#20
Hi, can anyone share Edexcel AS Further Maths Core Pure end of course test paper?
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