• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

Preliminary mathematics marathon (3 Viewers)

edmundsung

Member
Joined
Dec 16, 2009
Messages
31
Gender
Undisclosed
HSC
2012


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

Banned
Joined
Dec 15, 2008
Messages
609
Gender
Male
HSC
2009


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)
 
Last edited:

edmundsung

Member
Joined
Dec 16, 2009
Messages
31
Gender
Undisclosed
HSC
2012
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

Member
Joined
Mar 12, 2009
Messages
62
Gender
Male
HSC
2010


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

nat_doc

Member
Joined
May 17, 2010
Messages
82
Gender
Female
HSC
2010
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
 
Last edited:
Joined
Dec 31, 2009
Messages
4,741
Location
sarajevo
Gender
Female
HSC
2015
Uni Grad
2017
Re: another question

Much fun was had in this thread.

Quick, someone post a new question!
 

Users Who Are Viewing This Thread (Users: 0, Guests: 3)

Top