VCE Methods Questions Thread
find values of a and b such that a(x+2)+b(x+3)=18x+8 for all values of x
if someone could help that would be great
chemistry1111 Expand then factorising LHS gives (a+b)x + (2a + 3b) = 18x + 8 --> by comparing coefficients, you obtain a+b = 18 and 2a+3b =8, solve simultaneously for a and b, giving a=46 and b=-28
thank you so much that makes sense
also this question
For the polynomial P(x)=(a+1)x3 + (b-7)x2 +c+5, find values for a,b and c if P(x) has a degree 2, a leading coefficient of 3 and the constant term is -1
when factorizing polynomials in methods unit 3/4, what method do most people use vcaa prefer, long division or equating coefficients
Do we need to buy the new Methods 3/4 book for the new study design (2023)?
I feel like the textbook questions are the same...
Can anyone plz help with this question (part b, c and d in particular):
PizzaMaster
Howdy, I did QCE Methods, so if I mess this up, I’m really sorry.
For b) may I know what the answers are? I may be able to work backwards
I didn’t do d) because I wasn’t that confident in doing the question.
Hope this helps,
-jinx_58
Hi jinx_58,
Thank you so much for solving!
The answer to part b is Domain = [0, 6]; Range = [0, 9/2]
The answer to part c is 9/2
- Edited
The graph of y = 5/x -6 is reflected in the x-axis and
then in the y-axis. The equation of the final image is:
could someone help with this question
- Edited
- reflected in the x-axis
- reflected in the y-axis
**I'm not quite sure if y = (5/x) -6 or if it's y = 5/(x-6) so I'll show the working for both
let f[x] = (5/x) -6
x reflection: -f[x] = f1[x] = -[ (5/x) - 6 ] = -( 5/x ) + 6
y reflection: f1[-x] = f2[x] = -(5/ [-x] ) +6 = -(5/ x ) +6
let g[x] = 5/(x-6)
- x reflection: -g[x] = f1[x] = -[ 5/(x-6) ] = -5/(x-6)
- y reflection: g1[-x] = f2[x] = -5/( [-x] - 6) = -5/( -x - 6)
hope that helps!
Hi everyone,
Is (2x-1) the same as (x-1/2)
I mean they both give x= 1/2
But let's say if we had to factorise 12x3 + 20x2 − x − 6
and we get (2x-1)(2x+3)(3x+2)
Can I also write this as:
(x-1/2)(2x+3)(3x+2)?
ALSO can some please help and find the roots of this: 12x3 + 20x2 − x − 6
using rational root theorem
much appreciated!!!
- Edited
PizzaMaster
good question! to answer your first one, (2x-1) = 2 (x-1/2). they're not the same but the same x value will make both of them equal (0). I'm not home rn so I'll reply to the second one when I have access to my iPad.
EDIT
I think it's easier if you watch a video, here's a good one I found (watching 0:00-2:40 should be enough): https://www.youtube.com/watch?v=4XytYH35AP0
So I can write (2x-1)(2x+3)(3x+2) as 2(x-1/2)(2x+3)(3x+2) NOT (x-1/2)(2x+3)(3x+2)
Great, thanks crumblycupcakes!
With rational root theorem, I find that it is super time consuming... after we find the possible roots, is there a shortcut to know which ones to try first?
For example with 12x3 + 20x2 − x − 6:
the possible roots are +- [1/2, 1/3, 1/4, 1/6, 1/12, 2/3, 3/4, 3/2]
That is a lot! Is there a way to know which one we can try first?
Also does simplifying work?
for example with 12x3 + 20x2 − x − 6, we do like factors of 6/ factors of 12. Can we do factors of 1/ factors of 2 (since 6/12 = 1/2?)
show that 0=x2 + 8x +17 has no solutions
d= (8)2 -4(1)(17) =-4 is this the working out
Yep I think that should be enough but just to be safe and make sure I got the mark, I do something like
⁂ d<0 --> no real solutions
crumblycupcakes yeah you never know if VCAA will reject "no solutions" on the grounds that there are complex solutions