# P5 roots of equations

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#1
the roots of 5x^3 - 16x^2 - 12x + 3 = 0 are alpha, beta and gamma, what is the value of

alpha^3 + beta^3 + gamma^3?
0
15 years ago
#2
(Original post by fishpaste)
the roots of 5x^3 - 16x^2 - 12x + 3 = 0 are alpha, beta and gamma, what is the value of

alpha^3 + beta^3 + gamma^3?

can it be factorised? ...will try...!
0
#3
(Original post by ogs)
can it be factorised? ...will try...!
has to be done using the P5 methods I'm afraid
0
15 years ago
#4
I think it factorised to...

(5x-1)(x^2+3x-3)
0
15 years ago
#5
(Original post by fishpaste)
the roots of 5x^3 - 16x^2 - 12x + 3 = 0 are alpha, beta and gamma, what is the value of

alpha^3 + beta^3 + gamma^3?
The longhand way:

Rewrite the eq as x^3 - 16/5 x^2 - 12/5 x + 3/5 = 0.

Using theorems regarding the roots of equations:

a+b+c = 16/5
ab+bc+ac = -12/5
abc = -3/5.

Now consider (a+b+c)^3 - you can work out a^2+b^2+c^2 using the first two equations, and simplify to get a^3+b^3+c^3
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15 years ago
#6
alpha = 1/5
beta=(3+root21)/2
gamma=(3-root21)/2

therefore the sum of the cubes =54.0 3sf

i don't know how correct it is!
0
15 years ago
#7
The shorthand way is to factorise, although it's not easy to spot the factor 0
15 years ago
#8
a³ + b³ + c³ = (a + b + c)³ - 3(ab + bc + ac)(a + b + c) + 3abc
0
15 years ago
#9
there was an error on the factorising before! .. it really goes to..
(5x-1)(x^2-3x-3)...
p5 are you doing edexcel? cos we don't have to do this type of stuff...
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#10
Oh god, you guys I'm such a twit. I wrote down the wrong equation, the correct equation was

x^3 - 3x^2 + 5x - 7 = 0

I'm pretty sure they want me to do it by finding an expression for alpha^3 + beta^3 + gamma^3 in terms of the other things, but I'm not sure how to do this.
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#11
(Original post by It'sPhil...)
a³ + b³ + c³ = (a + b + c)³ - 3(ab + bc + ac)(a + b + c) + 3abc
This is the expression I wanted. Any tips of deriving these kind of expressions in the exam?

Thanks very much.

(this is OCR for teh person who asked)
0
15 years ago
#12
From the book you know that

ax^3 + bx^2 + cx + d=0

has roots and that

aS3 + bS2 + cS1=0

Where S1=alpha+beta+gamma

and S2=alpha^2 + beta^2 + gamma^2

etc.
0
15 years ago
#13
IN fact sorry

aS(3+r) + bS(2+r) + cS(1+r) + dS(r)=0

and r is an interger.
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#14
(Original post by PhilC)
IN fact sorry

aS(3+r) + bS(2+r) + cS(1+r) + dS(r)=0

and r is an interger.
Hm. I don't recognise that at all =o Not in the Cambridge OCR book per chance is it?
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15 years ago
#15
(Original post by fishpaste)
This is the expression I wanted. Any tips of deriving these kind of expressions in the exam?

Thanks very much.

(this is OCR for teh person who asked)
Im not sure. I got the expression from a STEP problem I did a while ago. I think the thing to do is expand (a + b + c)^n and see what you get, eg n = 2 is easy...

(a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc

so, a² + b² + c² = (a + b + c)² - 2(ab + ac + bc)

Also always bear in mind you know three things: a + b + c, ab + bc + ac, abc so any expression you derive must be in terms of these only
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15 years ago
#16
Yes, justr before Miscellenious Exercise 1
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#17
(Original post by PhilC)
Yes, justr before Miscellenious Exercise 1
Ah found it thank you very much!
0
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