# A level further maths help!!!

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#1
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#2
that picture wasn’t very clear
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1 year ago
#3
(Original post by Rae5)
that picture wasn’t very clear
Q1

The new equation will have roots and .

So, this means the new equation will be
...

agreed?

Your goal is to express and in terms of and 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 is just 1 so we don't even need to work on this one.

But for you want to first combine it into a single fraction and the denominator is already good for us. Just deal with the numerator... what's in terms of and ??

Same approach to all questions.
Last edited by RDKGames; 1 year ago
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1 year ago
#4
(Original post by Rae5)
that picture wasn’t very clear
If a quadratic has roots and then you can work in reverse. What I mean by this is for example has roots 4 and 1, we know this by factorising to get

So we can work in reverse, we know the brackets will be . 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.
Last edited by AdamCor; 1 year ago
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1 year ago
#5
If a quadratic has roots and then you can work in reverse. What I mean by this is for example has roots 4 and 1, we know this by factorising to get
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 . Although it's still possible to calculate etc. directly, it's going to be fairly tedious.

Instead, if we consider the new roots to be and , we need only find and to be able to write down the new quadratic.

Since we can rewrite to get , all we really need to do is the (hopefully familiar by now) calculation to find in terms of and . The calculation of 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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