finding the equation given the roots Watch

jakaloupe
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i thought i forgot how to do the "proper" method for part b so i'll reverse what you do to get the roots in the first place

z=2\pm 4i



z-2=\pm 4i



(z-2)^2=-16



(z-2)^2 +16=0



(z^2 -4z+4)+16=0



p=-4

q=20
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Zacken
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(Original post by jakaloupe)

i thought i forgot how to do the "proper" method for part b so i'll reverse what you do to get the roots in the first place

z=2\pm 4i



z-2=\pm 4i



(z-2)^2=-16



(z-2)^2 +16=0



(z^2 -4z+4)+16=0



p=-4

q=20
That's a perfectly fine method. The other is expanding (z - (2-4i))(z - (2 + 4i)) = z^2 - z(2-4i + 2 + 4i) + (2-4i)(2+4i)
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atsruser
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(Original post by jakaloupe)

i thought i forgot how to do the "proper" method for part b so i'll reverse what you do to get the roots in the first place

z=2\pm 4i



z-2=\pm 4i



(z-2)^2=-16



(z-2)^2 +16=0



(z^2 -4z+4)+16=0



p=-4

q=20
Looks fine, but the "proper" method would be the sum/product of roots formulae:

-p=\alpha +\beta = -\frac{b}{a}, q= \alpha \beta = \frac{c}{a}

which give the same results.
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jakaloupe
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(Original post by Zacken)
That's a perfectly fine method. The other is expanding (z - (2-4i))(z - (2 + 4i)) = z^2 - z(2-4i + 2 + 4i) + (2-4i)(2+4i)
I tried this but i think something went wrong.... >.>

(Original post by atsruser)
Looks fine, but the "proper" method would be the sum/product of roots formulae:

-p=\alpha +\beta = -\frac{b}{a}, q= \alpha \beta = \frac{c}{a}

which give the same results.
what is this? i've never seen this before :/ should i know this for fp1?
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RDKGames
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(Original post by jakaloupe)
what is this? i've never seen this before :/ should i know this for fp1?
I hope you're joking. How are you doing exam questions and not come across this? Of course you do, that's quite literally the first thing on the chapter as far as I'm aware.

An equation in the form ax^2+bx+c=0 has roots \alpha and \beta

When we consider the roots we get:
(x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x+\alpha \beta=0

Now when we consider our original quadratic we get:
x^2+\frac{b}{a}x+\frac{c}{a}=0

By comparison, we can see that for them to be the same:
\frac{b}{a}=-(\alpha + \beta) \rightarrow -\frac{b}{a}=\alpha + \beta

and

\frac{c}{a}=\alpha \beta
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The_Big_E
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The sum and product of roots appears on the IAL specification for Edexcel Further Mathematics AS (F1), but not for the Edexcel GCE Further Mathematics AS (FP1). Not sure about other boards.
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HapaxOromenon3
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(Original post by The_Big_E)
The sum and product of roots appears on the IAL specification for Edexcel Further Mathematics AS (F1), but not for the Edexcel GCE Further Mathematics AS (FP1). Not sure about other boards.
It is on FP1 for all boards, including Edexcel (UK) FP1.
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The_Big_E
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(Original post by HapaxOromenon3)
It is on FP1 for all boards, including Edexcel (UK) FP1.
I'm afraid not. Here are the first two chapters from the IAL F1 specification.

Name:  Screen Shot 2016-08-30 at 1.36.51 PM.png
Views: 116
Size:  279.9 KB

The second chapter is not anywhere to be found on the FP1 GCE specification as shown below.

Attachment 576566576568

And I can confirm after doing all FP1 papers and F1 papers, nothing about \alpha and \beta roots has ever come up in an FP1 paper, only F1.

Can't say about other boards though.
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HapaxOromenon3
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(Original post by The_Big_E)
I'm afraid not. Here are the first two chapters from the IAL F1 specification.

Name:  Screen Shot 2016-08-30 at 1.36.51 PM.png
Views: 116
Size:  279.9 KB

The second chapter is not anywhere to be found on the FP1 GCE specification as shown below.

Attachment 576566576568

And I can confirm after doing all FP1 papers and F1 papers, nothing about \alpha and \beta roots has ever come up in an FP1 paper, only F1.

Can't say about other boards though.
It's certainly on AQA, OCR (MEI and non-MEI), and CIE, so it seems strange that Edexcel chose not to include it. I guess it's all part of the "dumbing down" process...
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The_Big_E
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Could be, but very strange indeed.
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