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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: #1procrastinator on October 08, 2011, 10:45:39 am

Title: Methods 1+2 Questions
Post by: #1procrastinator on October 08, 2011, 10:45:39 am: Solve for x

$\frac{-x^2+5x-5}{x-2}$

Original equation is

$\frac{1}{x-2}+3-x=0$

I used the quadratic formula but and used the coefficients in the numerator and got close to the right answer, but I feel uneasy disregarding the denominator

What I got:

$\frac{5+\sqrt5}{2}$ and $\frac{5-\sqrt5}{2}$

The book's solution is something like:

$<br />\frac{-5-\sqrt5}{2}$ and $\frac{5+\sqrt5}{2}$

Thanks
Title: Re: Methods 1+2 Questions
Post by: pi on October 08, 2011, 10:54:11 am: $\frac{1}{x-2}+3-x=0$

$\frac{1}{x-2} = x - 3$

$1= x^2 - 5x +6$

$x^2 - 5x +5=0$

$x_{1,2} = \frac{5 \pm \sqrt{25 - 4(1)(5)}}{2}$

$x_{1,2} = \frac{5 \pm \sqrt{5}}{2}$

You're correct :)
Title: Re: Methods 1+2 Questions
Post by: #1procrastinator on October 08, 2011, 11:00:51 am: Could you explain the bit about ignoring the denominators please? Why doesn't it affect the coefficients?

Also, my TI-89 gave the same answers as the book so I wasn't sure
Title: Re: Methods 1+2 Questions
Post by: pi on October 08, 2011, 11:04:05 am: Quote from: #1procrastinator on October 08, 2011, 11:00:51 am
Could you explain the bit about ignoring the denominators please? Why doesn't it affect the coefficients?

What do you mean 'ignoring denominators'?

All the denominator of $\frac{-x^2+5x-5}{x-2}$ says (in terms of solutions) is that $x \ne 2$ for all of $x \in R$

edit: pretty sure your answer is correct, wolfram alpha says so as well
Title: Re: Methods 1+2 Questions
Post by: #1procrastinator on October 08, 2011, 11:32:55 am: That's all? I thought it meant divide by the function or something like that, thereby giving different coefficients

Oh ok, I might've entered something wrong...I'll recheck
Title: Re: Methods 1+2 Questions
Post by: #1procrastinator on October 19, 2011, 05:42:34 pm: ah yeah, I didn't notice the brackets

if you have an absolute value equation with absolute values on both sides, as in

|x+1| = 9|3-x|

In a video, the guy made both positive and then one negative and one positive. He didn't explain why, but from playing around with random equations, I'm guessing it's to find all cases - making both negative will always give positive right?

I'm not too comfortable with absolute values, is this covered in-depth in VCE math? or is it yr 8/9/10 math?

and when you have a complex fraction (i believe that's what i saw them being referred to as), such as [(1/3) + (1/9)]/3, why must you separate the fractions before flipping them?

i.e. you can't say it's equivalent to (1/3)/(3+9),

i know you have to to get the right answer but i can't yet deduce the reason why
Title: Re: Methods 1+2 Questions
Post by: #1procrastinator on October 28, 2011, 03:46:06 pm: This is from the extended response part of quadratics in the Essential book

Piece of wire (12cm) cut into two pieces. One used to form square shape, other a rectangular in which length is twice its width

If x is side length of square, write dimensions of rectangle nterms of x

I got (6-2x)/3 for width and (12-4x)/3 for length - answer given for width is (12-4x)/6

does it have to be that way or can you simplify it? and why would they not simplify it, is it just to show 12-4x?
Title: Re: Methods 1+2 Questions
Post by: dinosaur93 on October 28, 2011, 04:38:27 pm: Anyone have Unit 2 Methods Practise Exams?
Title: Re: Methods 1+2 Questions
Post by: Panicmode on October 29, 2011, 02:14:52 pm: Quote from: #1procrastinator on October 28, 2011, 03:46:06 pm
This is from the extended response part of quadratics in the Essential book

Piece of wire (12cm) cut into two pieces. One used to form square shape, other a rectangular in which length is twice its width

If x is side length of square, write dimensions of rectangle nterms of x

I got (6-2x)/3 for width and (12-4x)/3 for length - answer given for width is (12-4x)/6

does it have to be that way or can you simplify it? and why would they not simplify it, is it just to show 12-4x?

Your answer is correct, it's perfectly okay to simplify. I suspect the reason they left it unsimplified has got something to do finding area and making it easier.

ie. It's easier to say ; ((12-4x)^2)/(3x6) then it is to write ((6-2x)(12-4x))/(3x3)
Title: Re: Methods 1+2 Questions
Post by: #1procrastinator on October 31, 2011, 01:36:37 pm: ^ thanks, just wanted to make sure
Title: Re: Methods 1+2 Questions
Post by: #1procrastinator on November 02, 2011, 06:00:57 pm: how come using the finite differences method doesn't gives the wrong solutions when finding the equation to these plots

(0, 80) (2, 64) (4, 54) (6, 48) (8,44)

I got y = -1/3x^3 + 4x^2 - 19(2/3)x + 80 using the equations from the math quest book