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Preliminary mathematics marathon (1 Viewer)

edmundsung

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i know it's easy for u guys...
but i still don't know how to solve it by multiplying the square of the denominator.
greatly appreciated if someone can help :)
 

random-1005

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i know it's easy for u guys...
but i still don't know how to solve it by multiplying the square of the denominator.
greatly appreciated if someone can help :)

mmm hsc 2012, looking to get a head start?

First step: exclude values where denominator=0 { however, you will see this only matters when you have a "less than or equal" or a "greater than or equal"}

(2x+5)(x+1)^2 / (x+1) < 3(x+1)^2 { DONT EXPAND, it is easier to move all terms to one side and factorise, saves time}
(2x+5)(x+1)-3(x+1)^2 <0
(x+1) { (2x+5) -3(x+1) } <0
(x+1)( 2-x) <0
x<-1, x > 2 ( if you need explaination from the previous step to this one please ask)
 
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edmundsung

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mmm hsc 2012, looking to get a head start?

First step: exclude values where denominator=0 { however, you will see this only matters when you have a "less than or equal" or a "greater than or equal"}

(2x+5)(x+1)^2 / (x+1) < 3(x+1)^2 { DONT EXPAND, it is easier to move all terms to one side and factorise, saves time}
(2x+5)(x+1)-3(x+1)^2 <0
(x+1) { (2x+5) -3(x+1) } <0
(x+1)( 2-x) <0
x<-1, x > 2 ( if you need explaination from the previous step to this one please ask)
thank you so much:cake:
 

Rezen

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If thats not enough proof i can think of two other methods to derive the polynomial.
 
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nat_doc

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x^2 -q=0

sum or roots of new equation is 0 and product is: -(a-b)(a-b) =-(a-b)^2= -sqrt(P+q-4abc+2sin\theta) *using the guassian distribution formula*
then upon inspection, product of roots = -q thus x^2 -q=0
 
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Re: another question

Much fun was had in this thread.

Quick, someone post a new question!
 

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