VCE Methods Question Thread!

[Bhootnike]

Lol synthetic division is NOT allowed

Long division isn't that much harder/tedious than synthetic division anyway.

its a lot quicker.

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[paulsterio]

Honestly, the synthetic division/normal division, won't really matter in year 12, I use short division, which only works when you're dividing by a factor, but really, you won't get marked for it, you can avoid division altogether by equating co-efficients

[Hutchoo]

Lol synthetic division is NOT allowed

VCAA has a document that says synthetic is allowed...just found out from the other thread.

Link plz =X

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[Phy124]

Lol synthetic division is NOT allowed

VCAA has a document that says synthetic is allowed...just found out from the other thread.

Link plz =X

http://www.vcaa.vic.edu.au/vce/studies/mathematics/cas/mmcasfaqs2011.pdf

The method by which this is carried out may involve long
division, synthetic division, equating coefficients or a similar approach

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[Bhootnike]

Wonderful

[pi]

Looks like I owe you guys an apology

[Bhootnike]

Looks like I owe you guys an apology

mwahah.. 

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[Hutchoo]

Are all/most multiple choice questions tech-allowed?

[pi]

Are all/most multiple choice questions tech-allowed?

Yep, MC qs are only in Exam 2

[b^3]

Are all/most multiple choice questions tech-allowed?

The Multiple Choice questions are on Exam 2, which is the tech-allowed exam. But it also depends on your school for SACS, for example we did get a set of multiple-choice questions on a tech-free SAC once. Just be prepared but for exams, the multiple choice questions are tech-allowed.

[Phy124]

Are all/most multiple choice questions tech-allowed?

The Multiple Choice questions are on Exam 2, which is the tech-allowed exam. But it also depends on your school for SACS, for example we did get a set of multiple-choice questions on a tech-free SAC once. Just be prepared but for exams, the multiple choice questions are tech-allowed.

Yeah same for us in regards to SACs. Although, they were generally very simple or just there to trip up a few students to differentiate the top students from the middle-range on short SACs i.e. questions where students are inclined to answer without putting a lot of thought towards the answer.

e.g. the first question on our first SAC was a multiple choice which consisted of a cubic graph with x-intercept "a" (on the negative side of the x-axis) and a turning point/x-intercept at "b" (on the positive side of the x-axis).

Many students were inclined to choose y = (x+a)(x-b)², when the answer is (x-a)(x-b)²

Anyway... Hutchoo, if you come up against multiple choice questions in the book you should probably use a calculator because that's what they are written for. If you're not supposed to use a calculator there is usually a little symbol next the question stating so.

[karenn]

I don't understand finding equations/simultaneous equations through the use of parameter.
Why do you do it? and I understand the process of working it out, but I need to know why!
For example:
The system of equations
 x + y + z + w = 4,
x + 3y + 3z = 2
x + y + 2z − w = 6 has infinitely many solutions. Describe this family of solutions and give the unique solution
when w = 6.

Why would you firstly subtract equation 2 from equation 3?
Thanks in advance

[brightsky]

x+y+z=4-w....(1)
x+3y+3z=2....(2)
x+y+2z=6+w..(3)
(2)-(1):
2y+2z=w-2...(4)
(2)-(3):
2y+z=-w-4...(5)
(4)-(5):
z=2w+2
y=(-3w-6)/2
x=(10-3w)/2
now we say that the system of equations has infinitely many solutions because w can be any number. not sure what the question means by 'describe this family of solutions' but to find the unique solution when w = 6, simply sub in w=6 into the above.

also, to answer your question more directly, you don't have to subtract eqn 2 from eqn 3 first. however, you would do so because it's part of the 'elimination' method. your textbook should have a detailed discussion on this method, or, if not, google it.

[Special At Specialist]

4 variables requires 4 equations to solve for all variables.
Since you only have 3 equations, the best you can do is find the solution to 3 of the variables in terms of a fourth variable. But when they tell you that w = 6, you can then find that specific solution, since you already have the solution to x, y and z in terms of w.

[Hutchoo]

Are examiners harsh when it comes to factorising?
As in, do you have to say "Using (a-b)^3 = ...." before you factorise something, or can you simply write the answer?
Tbh, I don't really know how you can show steps for factorising lolol.