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October 08, 2025, 05:59:10 am

Author Topic: VCE Methods Question Thread!  (Read 5721581 times)  Share 

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eagles

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Re: VCE Methods Question Thread!
« Reply #3495 on: January 04, 2014, 06:12:54 pm »
+1
"x:" means "where x is"
Besides set difference, you can also use it to describe restrictive/maximal domains.

For instance:
A question such as "Find all the values for x where f(x)>4, where f(x)=x2"
An answer can be x: {x<-2 U x>2}


the 3/4 textbook has some notation i havent seen before can someone please clarify

when its talking about set difference of two sets A and B, and how its denoted A\B, it says

A\B={x:x is an element of A, x isnt an element of B }

what does the x: mean exactly? (you can sorta tell) and also when do i use it?
cheers

hobbitle

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Re: VCE Methods Question Thread!
« Reply #3496 on: January 04, 2014, 06:20:17 pm »
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So when ever we want to solve for x when there its a quadratic equation we must use the null factor law?

Goddamn high school is so weird... giving weird names to everything!  Fixed point iteration, what on earth....

Ummm equating the quadratic to zero basically gives you the opportunity to use all methods of solving.  Sorry, I don't know all of their VCE technical names.... the format ax^2 + bx + c = 0 just makes it easier to 'see' the solution or how to solve....
« Last Edit: January 04, 2014, 06:22:38 pm by hobbitle »
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Only Cheating Yourself

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Re: VCE Methods Question Thread!
« Reply #3497 on: January 04, 2014, 06:50:16 pm »
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I need help please

The temperature T ( C) in an office follows the equation T=5/8t^2-5t+30, where t is the time in hours the office climate control system has been operating.  How long does it take temperature to reach 20 C?

I first thought as t as x as this is easier for me as everyone is in a quadratic expression i use CTS by dividing everything by 5/8 perhaps this was my mistake?  i did 5/1 / 5/8 which =8 and 5/1 / 30/1=48
so
x^2-8/2^2-8/2^2=48
x^2-16-16+48
(x-4)+32=0
x=-32+4?  Am i doing something wrong
x=26? WHat?
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hobbitle

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Re: VCE Methods Question Thread!
« Reply #3498 on: January 04, 2014, 07:05:13 pm »
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I'll try to do it as much as possible in the same way you were doing it, but simpler.
As you know, we need to find t (or x, in your case) when T=20.

So,
20 = 5/8t^2 - 5t + 30
subtract 20 from both sides
0 = 5/8t^2 - 5t + 10
divide everything (like you wanted to do) by 5/8
0 = t^2 - 8t + 16
by inspection (i think you call it...?)
0 = (t - 4)(t - 4)
so t = 4
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Only Cheating Yourself

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Re: VCE Methods Question Thread!
« Reply #3499 on: January 04, 2014, 07:10:43 pm »
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I'll try to do it as much as possible in the same way you were doing it, but simpler.
As you know, we need to find t (or x, in your case) when T=20.

So,
20 = 5/8t^2 - 5t + 30
subtract 20 from both sides
0 = 5/8t^2 - 5t + 10
divide everything (like you wanted to do) by 5/8
0 = t^2 - 8t + 16
by inspection (i think you call it...?)
0 = (t - 4)(t - 4)
so t = 4

Yea its inspection…  Why do we minus both sides by 20?  Is it to get rid of T?  which is the subject?  but 20 C and 30 aren't like terms so how can you minus them?  And how do we know 20 C is the subject?
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hobbitle

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Re: VCE Methods Question Thread!
« Reply #3500 on: January 04, 2014, 07:17:49 pm »
+1
They are like terms... they are just numbers... its important to keep an eye on your units in regards to things like length (mm, cm, m) or volume etc..... but temperature is just a number.... and the equation to describe the change in T has no units....

Read the question again and try to interpret it logically.  You are given an equation that determines the temperature of the room t hours after it is turned on. 
I think of it in my head as a sentence.  "When the room is 20 degrees, the equation for the temperature in the room must equal 20.". 
You're given the equation for the room... so write it down.  Then make it equal 20.

You minus both sides by 20 so that you can equate the quadratic to zero which makes solving easier.  It's not really about 'getting rid of T'... it's about getting the equation in a form that you can solve.

That kind of becomes an important concept especially when dealing with worded questions down the track.  Your first steps are always to get the information you are given into a format that you know how to solve.  Rearranging, substituting, etc.
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SunnyB

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Re: VCE Methods Question Thread!
« Reply #3501 on: January 04, 2014, 07:50:42 pm »
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How do you facetorise x^2+4x+8? I know how u do it but my belief thinks the way I do it I'd wrong!!

Only Cheating Yourself

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Re: VCE Methods Question Thread!
« Reply #3502 on: January 04, 2014, 08:24:36 pm »
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How do you facetorise x^2+4x+8? I know how u do it but my belief thinks the way I do it I'd wrong!!

x^2+4-4+8
[(x^2+4)^2+4]
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Only Cheating Yourself

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Re: VCE Methods Question Thread!
« Reply #3503 on: January 04, 2014, 08:27:41 pm »
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They are like terms... they are just numbers... its important to keep an eye on your units in regards to things like length (mm, cm, m) or volume etc..... but temperature is just a number.... and the equation to describe the change in T has no units....

Read the question again and try to interpret it logically.  You are given an equation that determines the temperature of the room t hours after it is turned on. 
I think of it in my head as a sentence.  "When the room is 20 degrees, the equation for the temperature in the room must equal 20.". 
You're given the equation for the room... so write it down.  Then make it equal 20.

You minus both sides by 20 so that you can equate the quadratic to zero which makes solving easier.  It's not really about 'getting rid of T'... it's about getting the equation in a form that you can solve.

That kind of becomes an important concept especially when dealing with worded questions down the track.  Your first steps are always to get the information you are given into a format that you know how to solve.  Rearranging, substituting, etc.

I need a lot of help on worded questions!  Thanks for the help.
'My belief is stronger than your doubt'

BubbleWrapMan

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Re: VCE Methods Question Thread!
« Reply #3504 on: January 04, 2014, 08:28:06 pm »
+1
How do you facetorise x^2+4x+8? I know how u do it but my belief thinks the way I do it I'd wrong!!

That can't be factorised over the real numbers.
Tim Koussas -- Co-author of ExamPro Mathematical Methods and Specialist Mathematics Study Guides, editor for the Further Mathematics Study Guide.

Current PhD student at La Trobe University.

Only Cheating Yourself

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Re: VCE Methods Question Thread!
« Reply #3505 on: January 04, 2014, 08:30:18 pm »
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That can't be factorised over the real numbers.

Use CTS, but cause the last number is a positive you can't use DOTS so you can just leave it…
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hobbitle

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Re: VCE Methods Question Thread!
« Reply #3506 on: January 04, 2014, 08:31:07 pm »
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x^2+4-4+8
[(x^2+4)^2+4]

Mate - think about what you've done here (I can't even work out the logic actually). You lost the whole 'x' term. Where did it go? You can't just get rid of it.

As poster above says, it can't be factorised over the real numbers.
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hobbitle

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Re: VCE Methods Question Thread!
« Reply #3507 on: January 04, 2014, 08:32:00 pm »
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My Belief - you have an obsession with CTS but you also seem to be lacking confidence in using it. Maybe watch a Khan Academy video on CTSing?
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Re: VCE Methods Question Thread!
« Reply #3508 on: January 04, 2014, 08:33:42 pm »
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Mate - think about what you've done here (I can't even work out the logic actually). You lost the whole 'x' term. Where did it go? You can't just get rid of it.

As poster above says, it can't be factorised over the real numbers.

x^2+4x+8
x^2+4x+4-4+8
x^2+4x+4
{x+2)^2+4}
'My belief is stronger than your doubt'

SunnyB

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Re: VCE Methods Question Thread!
« Reply #3509 on: January 04, 2014, 08:37:40 pm »
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My belief is stronger then yours, that is what I sent you, you told me messages you couldldt figure it out and your the one who gave me the question.

You could complete the square? Use imaginary numbers?