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2005 Paper 2 Q11(b) - HSN forum

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2005 Paper 2 Q11(b)


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#1 Adam

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Posted 12 May 2006 - 06:27 PM

Okay, I got x^2 + (p-1)x + 1 as the quotient, but according to these solutions I'm looking at, it should be x^2 - (p-1)x + 1

Can somebody explain why it's - and not +?

Cheers.

#2 Michael

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Posted 12 May 2006 - 07:02 PM

Sorry Im just about to run so I dont have time to work through the question but you can get worked solutions here:

http://www.mathsrevision.com/index_files/M...2005_paper2.pdf

Mike

#3 Adam

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Posted 12 May 2006 - 07:07 PM

Interesting... this guide must be wrong then.

Looks like I was right afterall. tongue.gif

#4 John

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Posted 13 May 2006 - 08:36 PM

For part b) i get

(x+1)(x^2 - (1-p)x + 1)=0

so therefore i would use x^2 - (1-p) + 1 in the discriminant formula to find the range of values for which p gives the original equation Real and Distinct roots

a = 1
b = -(p-1)
c = 1

b^2 - 4ac = 0
(-(p-1))^2 - 4(1)(1) = 0
(-p+1)^2 - 4 = 0
p^2 - p - p + 1 - 4 = 0
p^2 - 2p - 3 = 0
(p - 3)(p + 1)

so there for -1 <= p => 3





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