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March 29, 2024, 06:59:31 pm

Author Topic: VCE Methods Question Thread!  (Read 4803261 times)  Share 

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Corey King

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Re: VCE Methods Question Thread!
« Reply #18870 on: October 17, 2020, 03:03:51 pm »
0
Still not sure how to get the answer through pen and paper. :/

ArtyDreams

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Re: VCE Methods Question Thread!
« Reply #18871 on: October 17, 2020, 03:30:50 pm »
+5
Heyy! I might be able to help with this.

So we have the expression: 2a2 + 5a - 3
So the first thing I do is multiply 2 by -3 (first and last coefficients of the equation)
This gives me minus 6. Now I look at the second term, and find 2 numbers that when I multiply I get -6 and add to get 5. These numbers in this case will be -1 and and 6.
I then substitute these numbers in place of 5a, like this.

2a2 - a + 6a - 3

After I form this expression, I factorise by grouping.

this gives me a(2a-1) + 3(2a -1)
which simplifies to (a+3)(2a-1).

I then just substitute the a term back in which is x-1
=(x-1+3)(2(x-1)-1)
=(x+2)(2x-3)

Thats it! Your CAS has taken out the number 2 from the second bracket.

I hope this makes somewhat sense! Someone else might be able to give you a better explanation  :)

EDIT: sorry I didn't see your post above when you tried using this method! I hope this makes a bit of sense though. I don't really know how to fully explain why it works, so hopefully someone else can help you out with that.
« Last Edit: October 17, 2020, 03:34:26 pm by ArtyDreams »

Corey King

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Re: VCE Methods Question Thread!
« Reply #18872 on: October 17, 2020, 03:56:16 pm »
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Heyy! I might be able to help with this.

So we have the expression: 2a2 + 5a - 3
So the first thing I do is multiply 2 by -3 (first and last coefficients of the equation)
This gives me minus 6. Now I look at the second term, and find 2 numbers that when I multiply I get -6 and add to get 5. These numbers in this case will be -1 and and 6.
I then substitute these numbers in place of 5a, like this.

2a2 - a + 6a - 3

After I form this expression, I factorise by grouping.

this gives me a(2a-1) + 3(2a -1)
which simplifies to (a+3)(2a-1).

I then just substitute the a term back in which is x-1
=(x-1+3)(2(x-1)-1)
=(x+2)(2x-3)

Thats it! Your CAS has taken out the number 2 from the second bracket.

I hope this makes somewhat sense! Someone else might be able to give you a better explanation  :)

EDIT: sorry I didn't see your post above when you tried using this method! I hope this makes a bit of sense though. I don't really know how to fully explain why it works, so hopefully someone else can help you out with that.

This was helpful thank you :) I can see the process of how it is done now.

Ah fair enough. My intuition tells me:
If you have a factorized equation: (x-1)(x-1)
Then you MULTIPLY to expand: x^2-2x+1
And if you want to find the factors of the equation, you DIVIDE (in finding the factors of the terms)
-1 x -1 = 1
-1 + -1 = -2
When trying to solve the problem I was trying to do this method :)
I know how the method you showed works now, but do not understand how it works.
You multiplied an expanded expression to find the factors (it looks like you are MULTIPLYING in order to DIVIDE to me, which I don't understand :P )

keltingmeith

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Re: VCE Methods Question Thread!
« Reply #18873 on: October 17, 2020, 04:40:31 pm »
+5
This was helpful thank you :) I can see the process of how it is done now.

Ah fair enough. My intuition tells me:
If you have a factorized equation: (x-1)(x-1)
Then you MULTIPLY to expand: x^2-2x+1
And if you want to find the factors of the equation, you DIVIDE (in finding the factors of the terms)
-1 x -1 = 1
-1 + -1 = -2
When trying to solve the problem I was trying to do this method :)
I know how the method you showed works now, but do not understand how it works.
You multiplied an expanded expression to find the factors (it looks like you are MULTIPLYING in order to DIVIDE to me, which I don't understand :P )

You know how all your teachers laugh about mathematicians being lazy, and you just sit there thinking, "but this is way more complicated, how is this them being lazy??" (or at least that's what it was like for me lmfao) Well, the idea is this - as a mathematician, I know how to factorise something of the form:

ax(bx+c)+d(bx+c)

But I don't know how to factorise something generally of the form:

ax^2+bx+c

But what if I made the second one look like the first? I could come up with an entirely new method for the second one (for which, completing the square enters the equation), but why don't I just manipulate things so that the first equation looks like the second equation? That's where ArtyDreams explanation comes in. Sure, they're taking extra steps to get to something like the first form (or as you put it, "multiplying in order to divide"), but that's so they can make something easier to work with. It's kind of similar to when you first learned how to expand - nowadays, if you wanted to expand:



You'd probably do it in one go using FOIL and get:



But originally, what you'd do is expand the first bracket, and then the second, like so:



This is the how and why of it working, though not the mechanics. Let us know if there's still something you still don't understand

S_R_K

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Re: VCE Methods Question Thread!
« Reply #18874 on: October 18, 2020, 11:39:44 am »
+4
Just to try to distill the explanation of why the "find factors of ac that sum to b" method works:

Consider \((px+q)(rx+s) = prx^2 + (ps + qr)x + qs\)

Note that \(ps\) and \(qr\) are factors of \(prqs\), so if we let If \(a = pr,\; b = ps + qr,\; c = qs\) then \(b\) is the sum of two factors of \(ac\).

Corey King

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Re: VCE Methods Question Thread!
« Reply #18875 on: October 18, 2020, 01:07:00 pm »
+1
All your responses added something, thanks guys :)
SDK helped me understand how the process works, and Kelting taught me why we use it.
Best forum ever :P

Coolgalbornin03Lo

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Re: VCE Methods Question Thread!
« Reply #18876 on: October 18, 2020, 02:06:45 pm »
+1
How do transformations effect integrals?

For example if the integral from 4 to 3 for f(x)= 6, how can we use this information to work out the integral from 4 to 3 for (5-2f(x))?

I ended up getting 5-12 which is 7 but the answer is three  :P. What I did was remove the 5 and 2 out of the integral.
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Corey King

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Re: VCE Methods Question Thread!
« Reply #18877 on: October 18, 2020, 03:45:18 pm »
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Hey guys :)
When I go to draw graphs, I usually do them in a 1-1 ratio.
So the distance from the origin for x=1 and y=1 is the same, on paper. and x=2 y=2 and so  on.
In the textbooks, the ratios are often different to this depending on what line the graph is representing.

For example, this graph on parabolas: https://gyazo.com/bef946857e0739a46bf738edf2aaa981
And this graph on parabolas:
https://gyazo.com/e987f47353d8efc386476b13a09c89ce

Is there a method / are there methods behind determining the scale on both axes when drawing a coordinate plane?

james.358

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Re: VCE Methods Question Thread!
« Reply #18878 on: October 18, 2020, 07:56:28 pm »
+5
How do transformations effect integrals?

For example if the integral from 4 to 3 for f(x)= 6, how can we use this information to work out the integral from 4 to 3 for (5-2f(x))?

I ended up getting 5-12 which is 7 but the answer is three  :P. What I did was remove the 5 and 2 out of the integral.

Hey Coolgalbornin03Lo!

This question doesn't really test your transformation skills. Rather, you have to know how to use integral properties to break down the integral to different "known components".

So if the question you provided is correct, then yes the answer would be 7. Some practice exam companies do have quite a bit of errors  ::)

Hey guys :)
When I go to draw graphs, I usually do them in a 1-1 ratio.
So the distance from the origin for x=1 and y=1 is the same, on paper. and x=2 y=2 and so  on.
In the textbooks, the ratios are often different to this depending on what line the graph is representing.

For example, this graph on parabolas: https://gyazo.com/bef946857e0739a46bf738edf2aaa981
And this graph on parabolas:
https://gyazo.com/e987f47353d8efc386476b13a09c89ce

Is there a method / are there methods behind determining the scale on both axes when drawing a coordinate plane?

Hey Corey!

Provided that you do label & provide the scale, the ratio between the x and y axis doesn't matter. In Maths Methods, when you draw the graph as long as you have correctly labeled the intercepts, endpoints and asymptotes with the correct shape, then you will always be awarded the marks even if the scale is a bit off. However, you do sometimes have to be aware that in some SACs, teachers will ask you to provide a scale. In that case, you can choose any scale as long as it is consistent.

Hope this helps!
James
« Last Edit: October 18, 2020, 07:59:54 pm by james.lhr »
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Corey King

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Re: VCE Methods Question Thread!
« Reply #18879 on: October 19, 2020, 11:16:56 am »
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Hey Coolgalbornin03Lo!

This question doesn't really test your transformation skills. Rather, you have to know how to use integral properties to break down the integral to different "known components".

So if the question you provided is correct, then yes the answer would be 7. Some practice exam companies do have quite a bit of errors  ::)

Hey Corey!

Provided that you do label & provide the scale, the ratio between the x and y axis doesn't matter. In Maths Methods, when you draw the graph as long as you have correctly labeled the intercepts, endpoints and asymptotes with the correct shape, then you will always be awarded the marks even if the scale is a bit off. However, you do sometimes have to be aware that in some SACs, teachers will ask you to provide a scale. In that case, you can choose any scale as long as it is consistent.

Hope this helps!
James

Very helpful James thank you :)
You mentioned having the correct shape. If the scale is off, this affects the shape. Is the cause of change in shape going to lose marks?

keltingmeith

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Re: VCE Methods Question Thread!
« Reply #18880 on: October 19, 2020, 11:38:27 am »
+5
Very helpful James thank you :)
You mentioned having the correct shape. If the scale is off, this affects the shape. Is the cause of change in shape going to lose marks?

As long as it's roughly fine, the markers won't care. They know you're not a machine which calculates the y-value for every time x changes by 0.00000001 to get the EXACT right shape. They know you'll make slight mistakes. As long as the general shape matches, y'all be gucci

Corey King

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Re: VCE Methods Question Thread!
« Reply #18881 on: October 19, 2020, 12:11:23 pm »
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As long as it's roughly fine, the markers won't care. They know you're not a machine which calculates the y-value for every time x changes by 0.00000001 to get the EXACT right shape. They know you'll make slight mistakes. As long as the general shape matches, y'all be gucci

Sweet :)

Corey King

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Re: VCE Methods Question Thread!
« Reply #18882 on: October 22, 2020, 02:33:44 pm »
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Hey guys,
I'm doing work on completing the square.
There is one small thing I don't understand with the kind of question I have attached. (solve for x)
When you don't get an (x+b)^2=0 expression but instead get an (x+b)^2=a expression,
the textbook will give the solution to x as x = -b+(root)a
I guess they want to just do x+b inside the brackets, be left over with (root)a and then square (root)a to have a=a.
Is this legal? When expanding a binomial multiplied by a binomial, don't you have to cross multiply?
Won't this equation be left with a 2bx to imbalance the equation?
Many thanks,
Corey

keltingmeith

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Re: VCE Methods Question Thread!
« Reply #18883 on: October 22, 2020, 04:02:46 pm »
+5
Hey guys,
I'm doing work on completing the square.
There is one small thing I don't understand with the kind of question I have attached. (solve for x)
When you don't get an (x+b)^2=0 expression but instead get an (x+b)^2=a expression,
the textbook will give the solution to x as x = -b+(root)a
I guess they want to just do x+b inside the brackets, be left over with (root)a and then square (root)a to have a=a.
Is this legal? When expanding a binomial multiplied by a binomial, don't you have to cross multiply?
Won't this equation be left with a 2bx to imbalance the equation?
Many thanks,
Corey

I'm gonna be honest - I have no clue what you're asking. What I do know is, looking at your attachment, that's wrong.

Everything is fine up until line 4. When moving from line 3 to line 4, you've removed the square from the brackets. Where has it gone? Remember - things can't just disappear in maths, they have to move somewhere or you must change both sides of the equation. On top of that, when you moved that 17/4 to the other side, you forgot to change its sign to +17/4. To remove the square from the brackets (and this is what the book is doing), you should take the square root of both sides. This will cancel out the squaring of the (x+5/2)^2, and leave a square root on the other side. However, since you don't know if +sqrt(17)/2 or -sqrt(17)/2 led to (x-5/2)^2, you will need to leave the second square root as +/-, just as the book does. The plus/minus thing might sound complicated, so I will come back to it. First, here's how the working should look, for full clarity:




On the plus/minus thing, imagine your equation is x^2=4. This is real simple, right - you square a number, it equals 4. What is that number? Well, there are two numbers that become 4 when you square them - 2^2=4, because 2*2=4. However, (-2)^2 ALSO equals 4, because a negative times a negative makes a positive. Hence, (-2)^2=-2*-2=4. So, the solution to x^2=4 ISN'T x=2, it's actually x=-2 OR 2. We commonly write this as x=-2,2, or as x=plus/minus 2.

Some of that you may have already known, but again, had no clue what you were actually asking, so had to make some guesses. Let me know if I haven't solved your confusion, and maybe try rewording what you're having trouble understanding?

Corey King

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Re: VCE Methods Question Thread!
« Reply #18884 on: October 22, 2020, 04:19:12 pm »
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I'm gonna be honest - I have no clue what you're asking. What I do know is, looking at your attachment, that's wrong.

Everything is fine up until line 4. When moving from line 3 to line 4, you've removed the square from the brackets. Where has it gone? Remember - things can't just disappear in maths, they have to move somewhere or you must change both sides of the equation. On top of that, when you moved that 17/4 to the other side, you forgot to change its sign to +17/4. To remove the square from the brackets (and this is what the book is doing), you should take the square root of both sides. This will cancel out the squaring of the (x+5/2)^2, and leave a square root on the other side. However, since you don't know if +sqrt(17)/2 or -sqrt(17)/2 led to (x-5/2)^2, you will need to leave the second square root as +/-, just as the book does. The plus/minus thing might sound complicated, so I will come back to it. First, here's how the working should look, for full clarity:




On the plus/minus thing, imagine your equation is x^2=4. This is real simple, right - you square a number, it equals 4. What is that number? Well, there are two numbers that become 4 when you square them - 2^2=4, because 2*2=4. However, (-2)^2 ALSO equals 4, because a negative times a negative makes a positive. Hence, (-2)^2=-2*-2=4. So, the solution to x^2=4 ISN'T x=2, it's actually x=-2 OR 2. We commonly write this as x=-2,2, or as x=plus/minus 2.

Some of that you may have already known, but again, had no clue what you were actually asking, so had to make some guesses. Let me know if I haven't solved your confusion, and maybe try rewording what you're having trouble understanding?

It looks like I rushed transcribing the working to a new sheet of paper :P
Whoops.

You did answer my question :), as I completely forgot that I could take the square root of a binomial expression.

So I thought they were maybe doing something funky by substituting in a value for x into step 3 of your working. But since x does not equal zero, they were adding onto the value of x to also get (25/4) to the expanded value of the binomial.

Hopefully that helps you see where my head was at. I was clearly lost :P

Thanks again Kelting :)