Rae5
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Please help me I am so confused with how to work out the new equations for each question anyone please help me
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Rae5
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RDKGames
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(Original post by Rae5)
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Q1

The new equation will have roots \dfrac{\alpha}{\beta} and \dfrac{\beta}{\alpha}.

So, this means the new equation will be
x^2 - \left( \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} \right)x + \dfrac{\alpha}{\beta} \dfrac{ \beta }{\alpha}...

agreed?


Your goal is to express \dfrac{\alpha}{\beta} + \dfrac{ \beta }{\alpha} and  \dfrac{\alpha}{\beta}\dfrac{ \beta }{\alpha} in terms of \alpha + \beta and \alpha \beta because you know those from part (a) and so you can just sub those numbers in and simplify the coefficients to nice numbers!

Obviously, the constant term  \dfrac{\alpha}{\beta} \dfrac{ \beta}{\alpha} is just 1 so we don't even need to work on this one.

But for  \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} you want to first combine it into a single fraction \dfrac{\alpha^2 + \beta^2}{\alpha \beta} and the denominator is already good for us. Just deal with the numerator... what's \alpha^2 + \beta^2 in terms of \alpha + \beta and \alpha \beta ??


Same approach to all questions.
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AdamCor
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(Original post by Rae5)
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If a quadratic has roots \alpha^{3}\beta+1 and \alpha\beta^{3}+1 then you can work in reverse. What I mean by this is for example x^2+5x+4=0 has roots 4 and 1, we know this by factorising to get (x+1)(x+4)=0

So we can work in reverse, we know the brackets will be (x+\alpha^{3}\beta+1)(x+\alpha \beta^{3}+1). So you can expand this out to get the equation.

You should probably stick your values for alpha and beta in before you expand, otherwise it'll get messy. Then once you expand these brackets you'll arrive at a quadratic that has those roots, and you know this because at the start you put these as the roots.
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DFranklin
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(Original post by AdamCor)
If a quadratic has roots \alpha^{3}\beta+1 and \alpha\beta^{3}+1 then you can work in reverse. What I mean by this is for example x^2+5x+4=0 has roots 4 and 1, we know this by factorising to get (x+1)(x+4)=0
Your example is somewhat unrealistic; it is the norm for the polynomials in these questions to be chosen so that the roots are "nasty". For example, for the actual polynomial in the question (x^2+2x-5), the roots are -1 \pm \sqrt{6}. Although it's still possible to calculate \alpha^3\beta + 1 etc. directly, it's going to be fairly tedious.

Instead, if we consider the new roots to be \gamma = \alpha^{3}\beta+1 and \delta = \alpha\beta^{3}+1, we need only find \gamma + \delta and \gamma\delta to be able to write down the new quadratic.

Since we can rewrite to get \gamma + \delta = \alpha\beta(\alpha^2+\beta^2) + 2, all we really need to do is the (hopefully familiar by now) calculation to find \alpha^2 + \beta^2 in terms of \alpha+\beta and \alpha\beta. The calculation of \gamma\delta is even easier (once you've found \gamma+\delta).

You should probably stick your values for alpha and beta in before you expand, otherwise it'll get messy. Then once you expand these brackets you'll arrive at a quadratic that has those roots, and you know this because at the start you put these as the roots.
Again, this is usually going to be one of the more unpleasant ways of solving the problem. [It also fails *hard* if you ever need to do a problem for a cubic / quartic where the roots don't come out easily].
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