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FP1 Alpha Beta roots

Paper: http://filestore.aqa.org.uk/subjects/AQA-MFP1-QP-JUN12.PDF
Question: 1)b)

I am highly surely I have done this question before with ease, but after coming back to it now I am struggling to make the jump from a+b and ab to what they write as a/b + b/a?

I have read the MS but could someone please step by step show me how to reach what they have asked to show.
Reply 1
Original post by PrinceUpsb


I am highly surely I have done this question before with ease, but after coming back to it now I am struggling to make the jump from a+b and ab to what they write as a/b + b/a?


1. αβ+βα=α2+β2αβ\frac{\alpha}{\beta}+\frac{\beta}{\alpha} = \frac{\alpha^2+\beta^2}{\alpha \beta}

2. (α+β)2=α2+2αβ+β2(\alpha+\beta)^2 = \alpha^2+2\alpha\beta+\beta^2
Reply 2
you have to combine the fraction to have a common denominator,

so multiply the first half with alpha

and the second half with beta so it would be equivalent to (a^2+b^2)/ab

which is the same as: ( (a+b)^2 -2(aB) )/ab

then put in the values for a+b and ab to get the answer
Reply 3
Original post by atsruser
1. αβ+βα=α2+β2αβ\frac{\alpha}{\beta}+\frac{\beta}{\alpha} = \frac{\alpha^2+\beta^2}{\alpha \beta}

2. (α+β)2=α2+2αβ+β2(\alpha+\beta)^2 = \alpha^2+2\alpha\beta+\beta^2


step 2 is incorrect as a^2+b^2 does not equal (a+b)^2.

It helps to remember these:

a^2+b^2 = (a+b)^2-2(ab)

a^3+b^3 = (a+b)^3-3(ab)(a+b)
Reply 4
Original post by iAkash
step 2 is incorrect as a^2+b^2 does not equal (a+b)^2.

It helps to remember these:

a^2+b^2 = (a+b)^2-2(ab)

a^3+b^3 = (a+b)^3-3(ab)(a+b)


Step 2 is not incorrect

There was no suggestion that a2+b2=(a+b)2a^2+b^2 = (a+b)^2

You use step 2 to re-write the numerator of step 1


It does not help to remember the rearrangements of standard expansions ... it is better to understand how to use them
Reply 5
Original post by iAkash
step 2 is incorrect


It's not a step: it's a hint


as a^2+b^2 does not equal (a+b)^2.


That is true a.e.
Reply 6
Original post by atsruser
It's not a step: it's a hint



That is true a.e.


My bad sorry, i thought that was a step :frown:
Reply 7
Original post by TenOfThem
Step 2 is not incorrect

There was no suggestion that a2+b2=(a+b)2a^2+b^2 = (a+b)^2

You use step 2 to re-write the numerator of step 1


It does not help to remember the rearrangements of standard expansions ... it is better to understand how to use them


Ofcourse it helps to remember the rearrangements, from the point of view of the original exam question it saves a lot of time. Im sure everyone can do the rearrangements but in an exam; but time has to be managed carefully :smile:
(edited 10 years ago)
Reply 8
Original post by iAkash
Ofcourse it helps to remember the rearrangements, from the point of view of the original exam question it saves a lot of time. Im sure everyone can do the rearrangements but in an exam; but time has to be managed carefully :smile:


No it doesn't

There is more than enough time in the exam

The brain power used for remembering pointless re-arrangement would be put to better use understanding

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