STEP - Question Help - Cambridge Maths

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Pliskin
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#1
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#1
Hi, I'm working through STEP orientated worksheets (not the actual past papers) in preparation for STEP I/II exams and I've came across a question in which I'm completely stumped and spent hours trying to reach a solution. Any help/solutions would be greatly appreciated, thanks a lot.
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ghostwalker
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#2
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(Original post by Pliskin)
...
Knowing that \alpha_1,\alpha_2,\alpha_3 are the three roots of the polynomial, what three equations can you write down about the relationships between the roots, based on the coefficients of the polynomial (I'm assuming you've covered roots of polynomials, at least for quadratics).

Then try and rewrite your expressions involving those forms. That will take care of the first two parts.

If you've not covered roots of polynomials, then:
Spoiler:
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Given your polynomial has three roots, write it as the product of three factors, expand, and compare coefficients.

And watch out for the coeff of x^3




For the final part, \alpha_1 is a root of the polynomial, so .... That's only a small hint, but hopefully may be enough.
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brianeverit
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(Original post by Pliskin)
Hi, I'm working through STEP orientated worksheets (not the actual past papers) in preparation for STEP I/II exams and I've came across a question in which I'm completely stumped and spent hours trying to reach a solution. Any help/solutions would be greatly appreciated, thanks a lot.
Image
Would these notes help?
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Pliskin
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(Original post by ghostwalker)
Knowing that \alpha_1,\alpha_2,\alpha_3 are the three roots of the polynomial, what three equations can you write down about the relationships between the roots, based on the coefficients of the polynomial (I'm assuming you've covered roots of polynomials, at least for quadratics).

Then try and rewrite your expressions involving those forms. That will take care of the first two parts.

If you've not covered roots of polynomials, then:
Spoiler:
Show

Given your polynomial has three roots, write it as the product of three factors, expand, and compare coefficients.

And watch out for the coeff of x^3




For the final part, \alpha_1 is a root of the polynomial, so .... That's only a small hint, but hopefully may be enough.
Hi, thanks a lot for replying. I'm a Scottish student so haven't covered A-level Further Maths, I found out that this is Fp1 content on quadratics/polynomials, so I found some helpful notes. I managed to get b) by expanding (a1+a2+a3)^2, but I'm struggling with a) and c) a bit still... I'm trying the same method for c) as b); expanding (a1+a2+a3)^2.
Any help is majorly appreciated, thanks a lot !!!
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Zenarthra
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#5
(Original post by Pliskin)
Hi, I'm working through STEP orientated worksheets (not the actual past papers) in preparation for STEP I/II exams and I've came across a question in which I'm completely stumped and spent hours trying to reach a solution. Any help/solutions would be greatly appreciated, thanks a lot.
Image

Alright i just had a go at the question:

So a1, a2 and a3 are roots.

Therefore: (x-a1)(x-a2)(x-a3) = 5x^3 - x^2 - 2x +3

Multiplying out left side gives:
x^3 - x^2(a1+a2+a3)+(a2a3 + a1a2 + a1a3) - a1a2a3

Since 1/a1 + 1/a2 + 1/a3 can be written as a1a2 + a2a3 + a1a3 / a1a2a3
Then when we let x = 0 on both sides we see a1a2a3 = -3
next equating coefficients of x from both sides
gives a2a3+a1a2+a1a3 = -2

Hence 1/a1 + 1/a2 + 1/a3 = -2/-3 = 2/3

Can you finish the rest?
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ghostwalker
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(Original post by Pliskin)
I managed to get b) by expanding (a1+a2+a3)^2
Yep, that's the standard method.

, but I'm struggling with a) and c) a bit still... I'm trying the same method for c) as b); expanding (a1+a2+a3)^2.
Any help is majorly appreciated, thanks a lot !!!
For a) put the LHS over a common denominator.

For c) I'd go with

since \alpha_1 is a root of the polynomial, then we know

5\alpha_1^3-\alpha_1^2-2\alpha_1+3=0

Similarly for the other two roots.

Now add your three equations.
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KongShou
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#7
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#7
The joy of Vieta Jumping

http://en.wikipedia.org/wiki/Vieta's_formulas
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Pliskin
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#8
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(Original post by Zenarthra)
Alright i just had a go at the question:

So a1, a2 and a3 are roots.

Therefore: (x-a1)(x-a2)(x-a3) = 5x^3 - x^2 - 2x +3

Multiplying out left side gives:
x^3 - x^2(a1+a2+a3)+(a2a3 + a1a2 + a1a3) - a1a2a3

Since 1/a1 + 1/a2 + 1/a3 can be written as a1a2 + a2a3 + a1a3 / a1a2a3
Then when we let x = 0 on both sides we see a1a2a3 = -3
next equating coefficients of x from both sides
gives a2a3+a1a2+a1a3 = -2

Hence 1/a1 + 1/a2 + 1/a3 = -2/-3 = 2/3

Can you finish the rest?
Thanks a lot for the help!! Greatly appreciated. I tried for c), the way that ghostwalker said to do it

"For c) I'd go with

since Image is a root of the polynomial, then we know

Image

Similarly for the other two roots.

Now add your three equations."

But I can't seem to get a solution from that method :confused:
Can you perhaps share how you would do it? Thanks a lot.
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davros
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#9
(Original post by Pliskin)
Thanks a lot for the help!! Greatly appreciated. I tried for c), the way that ghostwalker said to do it

"For c) I'd go with

since Image is a root of the polynomial, then we know

Image

Similarly for the other two roots.

Now add your three equations."

But I can't seem to get a solution from that method :confused:
Can you perhaps share how you would do it? Thanks a lot.
Are you happy that this method will give you 5\alpha_1^3 + 5\alpha_2^3 + 5\alpha_3^3 in terms of sums of lower powers of alpha?

Part (ii) gave you the sums of squares of the alphas and you should know (or be able to work out) the sums of the alphas themselves in terms of the original coefficients.
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Zenarthra
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#10
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#10
(Original post by Pliskin)
Thanks a lot for the help!! Greatly appreciated. I tried for c), the way that ghostwalker said to do it

"For c) I'd go with

since Image is a root of the polynomial, then we know

Image

Similarly for the other two roots.

Now add your three equations."

But I can't seem to get a solution from that method :confused:
Can you perhaps share how you would do it? Thanks a lot.
Luke sky walkers method was to add 3 equations together:
Inputing the individual roots and add 3 equations together gives;
5(a1^3 + a2^3 + a3^3)-2(a1+a2+a3)-(a1^2+a2^2+a3^2)+9=0

You can get a1+a2+a3 from equating the coefficients but remember to divide by 5.
(a1^3 + a2^3 + a3^3) is what you want, (a1^2+a2^2+a3^2) is what you already found.
Therefore you can solve c)
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Thereisnospoon!
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#11
were did you get these notes?
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TeeEm
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#12
(Original post by Thereisnospoon!)
were did you get these notes?
what notes?
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Thereisnospoon!
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#13
(Original post by TeeEm)
what notes?
the OPs notes
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TeeEm
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(Original post by Thereisnospoon!)
the OPs notes
probably referring to 2 PDF booklets for practicing step.
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WillWalker23
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#15
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#15
By using roots of polynomial laws:

alpha + beta + gamma = -b/a
alpha.beta + alpha.gamma + gamma.beta = c/a
alpha.beta.gamma = -d/a

When you get, 1/alpha + 1/beta + 1/gamma, you can merge the fractions to form (alpha.beta + alpha.gamma + beta.gamma)/(alpha.beta.gamma), this will help with the solution.

Use a similar approach for the second and third questions. E.g
Alpha^2 + Beta^2 + Gamma^2 can be thought of as
(Alpha + Beta + Gamma)^2 - 2(alpha.beta + beta.gamma and alpha.gamma), use this idea for the third question as well.
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