Roots of equations

Soldato
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Harro everyone

The textbook doesn't help. We haven't covered it in class. No one is online on facebook that can help.

OcUK is my last resort :o

The equation z^3 + 4z^2 - 3z + 1 = 0 has roots a, b and c

Find the values of:

a + b + c
ab + bc + ac
abc


Somehow I need to factorise the equation, cos then I'll get values for -a, -b and -c. But how? This question comes up in every exam paper (further pure 1) and I've got no idea how to do it

Thanks friends :)
 
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That's a horrible question.

Usually there's an obvious root say -1 or 1 which you can use.

When there's an obvious root to be found, say a. You can divide the cubic by (x - a) and then factorise the remaining quadratic.
 
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It's a horrible question in terms of having to factorise it. If there are identities that you can use to avoid factorising I guess it's not too bad. Been a while since I did FP1 though
 
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Does seem pretty nasty. Usually you'd just put a number in for z until it equals zero, then you know it is a root and the factor is (x minus whatever the number is). Using that you can then create a quadratic factor, and factorise that to get the other two.

Edit: beaten..and realised it's FP1 not C1 :p
 
Soldato
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I think I've found the rules. No thanks to the textbook or my teacher :rolleyes:

a+b+c = -b/a
ab+bc+ac = c/a
abc = -d/a

For the equation ax^3 + bx^2 + cx + d

so a + b + c = -4/1 = -4
ab + bc + ac = -3/1 = -3
abc = -1/1 = -1

Them's the answers apparently. Still, I'm thinking wtf :o
 
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It's confusing in the way that you are using a, b, c as the roots of the equations as well as the coeffecients of each term.

think of:
ax^3 + bx^2 + cx + d = 0 having roots f, g and h:
then
f + g + h = -b/a
fg + gh + fh = c/a
fgh = -d/a
 
Soldato
OP
Joined
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It's confusing in the way that you are using a, b, c as the roots of the equations as well as the coeffecients of each term.

think of:
ax^3 + bx^2 + cx + d = 0 having roots f, g and h:
then
f + g + h = -b/a
fg + gh + fh = c/a
fgh = -d/a

Yeah that's true. It's only because it was alpha beta gamma initially, but that's very long winded to write.

If there's nothing in your textbook then bottom of page 8 and top of 9 of this may do some explaining, http://www.mathshelper.co.uk/OCR FP1 Revision Sheet.pdf


Wow thanks mate, they've listed all the identities there! :)
 
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When did they start doing rule based mathematics at A-level? How is that pure maths?

I hope they teach you to prove those rules.
 
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Associate
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When did they start doing rule based mathematics at A-level? How is that pure maths?

I hope they teach you to prove those rules.

Unfortunately on me exam board (OCR), there's seems to be very little of deriving things from first principles. They teach you in the book but it's never needed in the exam.
 
Soldato
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Harro everyone

The textbook doesn't help. We haven't covered it in class. No one is online on facebook that can help.

OcUK is my last resort :o

The equation z^3 + 4z^2 - 3z + 1 = 0 has roots a, b and c

Find the values of:

a + b + c
ab + bc + ac
abc


Somehow I need to factorise the equation, cos then I'll get values for -a, -b and -c. But how? This question comes up in every exam paper (further pure 1) and I've got no idea how to do it

Thanks friends :)

You've put me back on the drink again.
 
Last edited:
Associate
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When did they start doing rule based mathematics at A-level? How is that pure maths?
It's about algebraic identities, how is it not pure mathematics, in the A Level sense?

I hope they teach you to prove those rules.
It's trivial to prove. In fact you can prove the general result in precisely the same manner but due to lack of LaTeX support I'll do the cubic case.

f(x) = ax^3 + bx^2 + cx + d. Let the three roots be A, B, C, ie f(A) = f(B) = f(C) = 0. By factorisation theorems if f(A) = 0 then f(x) is exactly divisible by x-A and likewise for B and C. Thus f(x) is exactly divisible by (x-A)(x-B)(x-C). The coefficient of x^3 in that is 1 so we overall scale by a to get f(x) = a(x-A)(x-B)(x-C). This has all been done without doing actual expansions and factorisations. We expand this form of f(x) and equate the two formulae, f(x) = a(x-A)(x-B)(x-C) = ax^3 + a(A+B+C)x^2 + a(AB+BC+CA)x + a(ABC) = ax^3 + bx^2 + cx + d. Therefore b = a(A+B+C), c = a(AB+BC+CA), d = a(ABC). QED.

The generalise result for order N polynomials is done in a likewise manner.

Unfortunately on me exam board (OCR), there's seems to be very little of deriving things from first principles. They teach you in the book but it's never needed in the exam.
Nice signature, though it isn't gauge invariant :p
 
Soldato
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It's about algebraic identities, how is it not pure mathematics, in the A Level sense?

But simply using the identities given in the formula book shows no understanding? It's not even applying the factor theorem. At A-level you begin to understand everything from first principles.

It's trivial to prove. In fact you can prove the general result in precisely the same manner but due to lack of LaTeX support I'll do the cubic case.

f(x) = ax^3 + bx^2 + cx + d. Let the three roots be A, B, C, ie f(A) = f(B) = f(C) = 0. By factorisation theorems if f(A) = 0 then f(x) is exactly divisible by x-A and likewise for B and C. Thus f(x) is exactly divisible by (x-A)(x-B)(x-C). The coefficient of x^3 in that is 1 so we overall scale by a to get f(x) = a(x-A)(x-B)(x-C). This has all been done without doing actual expansions and factorisations. We expand this form of f(x) and equate the two formulae, f(x) = a(x-A)(x-B)(x-C) = ax^3 + a(A+B+C)x^2 + a(AB+BC+CA)x + a(ABC) = ax^3 + bx^2 + cx + d. Therefore b = a(A+B+C), c = a(AB+BC+CA), d = a(ABC). QED.

The generalise result for order N polynomials is done in a likewise manner.

Thanks. That is a question which requires understanding.
 
Last edited:
Soldato
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But simply using the identities given in the formula book shows no understanding? At A-level you begin to understand everything from first principles.

Thanks. That is a question which requires understanding.

No s*** :D Surely the bloody computer can work it out? lol !
 
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