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March 29, 2024, 10:55:14 pm

Author Topic: VCE Methods Question Thread!  (Read 4803443 times)  Share 

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darkz

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Re: VCE Methods Question Thread!
« Reply #17640 on: February 10, 2019, 11:48:23 am »
+1
Hello AN,
Just had some trouble solving this cubic;
(Apologies for syntax errors)
f:[-2,4] →R
f(x)=4x^3-8x^2-16x+32
I am told to graph f(x) so I factorized the equation to 4(x-2)^2(x-2), however I cannot graph this as the (x-2)^2 touches at the same point as the (x-2) intercept.
If anyone can tell me where I went wong that would be great.
Thanks.  :)

To factor the cubic, simply use the factor theorem, so sub in values until \(f(x)=0\)

\[
\begin{aligned}
f(x)&=4x^3-8x^2-16x+32\\
&=4(x^3-2x^2-4x+8)\\
&=4(x-2)^2(x+2)\\
\text{When }f(x)&=0\\
x&=-2,2\\
\end{aligned}
\]
« Last Edit: February 10, 2019, 11:49:57 am by darkz »
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C14M8S

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Re: VCE Methods Question Thread!
« Reply #17641 on: February 10, 2019, 12:11:20 pm »
0
Are there any quick methods to solve questions like the one that I've attached in relation to simultaneous equations having infinite/singular/no solutions, be it with or without CAS? I understand how they're solved by substitution, but it's a painfully slow method and I'd like to have another technique to use.
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darkz

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Re: VCE Methods Question Thread!
« Reply #17642 on: February 10, 2019, 12:15:06 pm »
+2
Are there any quick methods to solve questions like the one that I've attached in relation to simultaneous equations having infinite/singular/no solutions, be it with or without CAS? I understand how they're solved by substitution, but it's a painfully slow method and I'd like to have another technique to use.
Substitution?
The way I like to do it would be to simply make \(y\) the subject of each and then equate the gradients and y-intercepts. There is also another method involving ratios/matricies, but I'm not too familiar with that one.
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dream chaser

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Re: VCE Methods Question Thread!
« Reply #17643 on: February 10, 2019, 12:24:26 pm »
+2
Are there any quick methods to solve questions like the one that I've attached in relation to simultaneous equations having infinite/singular/no solutions, be it with or without CAS? I understand how they're solved by substitution, but it's a painfully slow method and I'd like to have another technique to use.

Hi C14M8S,

Here is how I would do it. First make the 2 equations in the form y=mx+c. Then when you do that, it makes it easier to solve it. As they are asking for unique solutions, unique solutions occur when the gradient are not the same. Therefore you would equate the gradients of each equation together. What ever values you get for k suggest that those values either create no solutions or an infinite number of solutions between the 2 equations. Therefore, k would be an element of all real numbers apart from the k values you obtain.

Hope this helps. :)

TheIllusion

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Re: VCE Methods Question Thread!
« Reply #17644 on: February 10, 2019, 02:48:39 pm »
0
To factor the cubic, simply use the factor theorem, so sub in values until \(f(x)=0\)

\[
\begin{aligned}
f(x)&=4x^3-8x^2-16x+32\\
&=4(x^3-2x^2-4x+8)\\
&=4(x-2)^2(x+2)\\
\text{When }f(x)&=0\\
x&=-2,2\\
\end{aligned}
\]
Thanks darkz! :)
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TheIllusion

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Re: VCE Methods Question Thread!
« Reply #17645 on: February 11, 2019, 07:23:42 pm »
0
I was just wondering if we needed to study on simultaneous equations with 3 variables, as it seems that our teacher has skipped the subject. Do I need to know about the simultaneous equations or not?
Thanks. :)
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darkz

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Re: VCE Methods Question Thread!
« Reply #17646 on: February 11, 2019, 08:42:37 pm »
0
I was just wondering if we needed to study on simultaneous equations with 3 variables, as it seems that our teacher has skipped the subject. Do I need to know about the simultaneous equations or not?
Thanks. :)

Well I'm going to assume that you have solved simultaneous equations with two unknowns. And therefore, you should still be able to solve simultaneous equations with three variables, only that you'd be manipulating three equations. I'd assume that your teacher has skipped it because they feel as though you can handle it, but of course, if you don't know how to approach solving it when there are three variables, feel free to ask them since I'm pretty sure it is a relevant skill which you require.
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aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17647 on: February 11, 2019, 09:06:56 pm »
0
Hey, just a question -
Find the angle, correct to two decimal places, that the lines joining the given points make with the positive direction of the x-axis: \(\left(c,b\right)\ and\ \left(b,c\right)\)
I know how to do it when there are numbers and when it is positive but this just does not make sense?
Can someone help? I'm up to =
\(\tan\theta=\frac{c-b}{b-c}\)
I'm not sure what to do next if anyone can help me? Thanks!
Modified this because I was just going to skip the question until I came to something more weird??
 Find the gradient of a straight line which is:
a) inclined at an angle of 120 degrees to the positive direction of the x-axis (first quadrant)
I wrote one question because I wanted to try the rest of them
« Last Edit: February 11, 2019, 09:10:51 pm by aspiringantelope »

lzxnl

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Re: VCE Methods Question Thread!
« Reply #17648 on: February 11, 2019, 09:13:31 pm »
+1
Hey, just a question -
Find the angle, correct to two decimal places, that the lines joining the given points make with the positive direction of the x-axis: \(\left(c,b\right)\ and\ \left(b,c\right)\)
I know how to do it when there are numbers and when it is positive but this just does not make sense?
Can someone help? I'm up to =
\(\tan\theta=\frac{c-b}{b-c}\)
I'm not sure what to do next if anyone can help me? Thanks!
Modified this because I was just going to skip the question until I came to something more weird??
 Find the gradient of a straight line which is:
a) inclined at an angle of 120 degrees to the positive direction of the x-axis (first quadrant)
I wrote one question because I wanted to try the rest of them

Are you aware that \(\frac{c-b}{b-c} = -1\)?

For the next one, use the fact that \(m = \tan(\theta)\).
« Last Edit: February 11, 2019, 09:26:02 pm by lzxnl »
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darkz

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Re: VCE Methods Question Thread!
« Reply #17649 on: February 11, 2019, 09:15:14 pm »
+1
Hey, just a question -
Find the angle, correct to two decimal places, that the lines joining the given points make with the positive direction of the x-axis: \(\left(c,b\right)\ and\ \left(b,c\right)\)
I know how to do it when there are numbers and when it is positive but this just does not make sense?
Can someone help? I'm up to =
\(\tan\theta=\frac{c-b}{b-c}\)
I'm not sure what to do next if anyone can help me? Thanks!

\[
\begin{aligned}
\tan(\theta)&=\frac{c-b}{b-c}\\
&=\frac{-(b-c)}{b-c}\\
&=-1\\
\text{ }\\
\therefore \theta&=\tan^{-1}(-1)\\
&=\frac{-\pi}{4}\\
&\approx -0.79^c\\
\end{aligned}\\
\]
« Last Edit: February 11, 2019, 09:17:12 pm by darkz »
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aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17650 on: February 11, 2019, 09:16:22 pm »
0
Are you aware that \(\frac{b-c}{c-b} = -1\)?

For the next one, use the fact that \(m = \tan(\theta)\).
Wait is there a formula for that to be -1? Or did you just try random numbers? How did you work that out?
If so
-1 = \(\tan(\theta)\)
I'm also unsure what this turns to now >.< cause I'm not familiar with rearranging trig functions
WAIT SORRY
I know now cause it was just the wrong way around so i didnt understand at that point
Modification
\(\tan\theta=-1\)
\(\tan^{-1}\left(-1\right)=\theta\)
= -45

Had a look at and the answer was 135 degrees, is that supposed to be correct cause 180-45 = 135 but i'm not sure if there is such a formula
« Last Edit: February 11, 2019, 09:20:22 pm by aspiringantelope »

darkz

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Re: VCE Methods Question Thread!
« Reply #17651 on: February 11, 2019, 09:27:59 pm »
+1
Wait is there a formula for that to be -1? Or did you just try random numbers? How did you work that out?
If so
-1 = \(\tan(\theta)\)
I'm also unsure what this turns to now >.< cause I'm not familiar with rearranging trig functions
WAIT SORRY
I know now cause it was just the wrong way around so i didnt understand at that point
Modification
\(\tan\theta=-1\)
\(\tan^{-1}\left(-1\right)=\theta\)
= -45

Had a look at and the answer was 135 degrees, is that supposed to be correct cause 180-45 = 135 but i'm not sure if there is such a formula

Well I never really fully understood these questions haha, but then I'd just assume that they they're looking for a positive answer. So then yeh, that'd be 135 (Refer to the attached diagram)

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aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17652 on: February 11, 2019, 09:32:29 pm »
0
Well I never really fully understood these questions haha, but then I'd just assume that they they're looking for a positive answer. So then yeh, that'd be 135 (Refer to the attached diagram)
Ok thanks! Just wondering where did the -(b-c)/b-c (I thought it was c-b/b-c as well o.O) (not b-c/b-c) (-) negative sign come from? the one in front of b-c on the numerator.
Thanks

MB_

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Re: VCE Methods Question Thread!
« Reply #17653 on: February 11, 2019, 09:37:13 pm »
+1
Ok thanks! Just wondering where did the -(b-c)/b-c (I thought it was c-b/b-c as well o.O) (not b-c/b-c) (-) negative sign come from? the one in front of b-c on the numerator.
Thanks
-(b-c) is equal to c-b it's just writing it as -b+c and taking out -1 to get -(b-c)
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aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17654 on: February 11, 2019, 09:49:42 pm »
0
-(b-c) is equal to c-b it's just writing it as -b+c and taking out -1 to get -(b-c)

Where did the -1 come from o.O?