VCE Methods Question Thread!

[Lear]

Absolute legend Jazzycab! Thank you

[Mattjbr2]

Can someone please show me why this method isn't working? Or, where my error is? How do you arrive at the correct inequalities using the null factor law (without trial and error/checking every x value)?
The green numbers are merely for reference.

[Sine]

Can someone please show me why this method isn't working? Or, where my error is? How do you arrive at the correct inequalities using the null factor law (without trial and error/checking every x value)?
The green numbers are merely for reference.

First consider the inequality as an equation by replacing the > or < with = and find a solution. Those will be some of the endpoints, then you will need to graph the function and just read of that for inequalities by using the intersections you have already solved for.
In your case the graphs would be y = x^2 + x and y = 0

[TheAspiringDoc]

I think your mistake is in line 8. You can’t just say either of the factors are greater than 0. Either BOTH are greater than 0 or both are less than 0 for y>0. They’re interlinked because they’re multiplied.

[Mattjbr2]

First consider the inequality as an equation by replacing the > or < with = and find a solution. Those will be some of the endpoints, then you will need to graph the function and just read of that for inequalities by using the intersections you have already solved for.
In your case the graphs would be y = x^2 + x and y = 0

Well, yes.. But how would you do it if you couldn't graph it? Using only the null factor law without trial and error and without plugging them back into the original inequality and without graphing anything. If someone threw an inequality at you and you had no idea how to graph it and didn't have time to check every answer, how would you solve it? I graphed it merely to show that my inequality is incorrect; my question is why is it incorrect?

I think your mistake is in line 8. You can’t just say either of the factors are greater than 0. Either BOTH are greater than 0 or both are less than 0 for y>0. They’re interlinked because they’re multiplied.

Yeah.. but the words "and" and "or" are used interchangeably when it comes to intercepts, as demonstrated in every answer in the textbook. E.g.: in y=x^2 y=4 means that x= -2 or 2, or you can say x= -2 and 2. Still, that "and/or" issue doesn't influence the problem I'm having in line 10.

[RuiAce]

Yeah.. but the words "and" and "or" are used interchangeably when it comes to intercepts, as demonstrated in every answer in the textbook. E.g.: in y=x^2 y=4 means that x= -2 or 2, or you can say x= -2 and 2. Still, that "and/or" issue doesn't influence the problem I'm having in line 10.

Actually, "and" never happens. Because if \(x=-2\) and \(2\), then you're saying \(x\) is two numbers at once.

We aren't logically correct when we say "and". The correct way of saying it is always or. It just so happens that you don't get penalised on semantics, or else 50% of the cohort would lose marks on something pathetic like wording when they clearly understand the maths itself; it doesn't change the fact that in reality they actually do matter.

To explain how inequalities involving not one-to-one functions work, the use of "and" and "or" correctly is required.

[Sine]

Yeah.. but the words "and" and "or" are used interchangeably when it comes to intercepts, as demonstrated in every answer in the textbook. E.g.: in y=x^2 y=4 means that x= -2 or 2, or you can say x= -2 and 2. Still, that "and/or" issue doesn't influence the problem I'm having in line 10.

Hmm there is a distinction between the words "and" and "or" if that is what you mean?
"and" would mean that the solutions are occuring at the same time
Probably best to give your answers as x = -2, 2 by using a comma

[Mattjbr2]

Actually, "and" never happens. Because if \(x=-2\) and \(2\), then you're saying \(x\) is two numbers at once.

We aren't logically correct when we say "and". The correct way of saying it is always or. It just so happens that you don't get penalised on semantics, or else 50% of the cohort would lose marks on something pathetic like wording when they clearly understand the maths itself; it doesn't change the fact that in reality they actually do matter.

To explain how inequalities involving not one-to-one functions work, the use of "and" and "or" correctly is required.

But x is two numbers at once in a one-to-many graph?
I've figured out how to solve the inequality thing. You assume both factors are greater than zero as well as less than zero and then you remove the redundancies. E.g.: x>2 and x>3 implies only x>3 as seen here: https://www.khanacademy.org/math/algebra-home/alg-quadratics/alg-quadratic-inequalities/v/quadratic-inequality-example at the 5 minute mark

[RuiAce]

But x is two numbers at once in a one-to-many graph?
I've figured out how to solve the inequality thing. You assume both factors are greater than zero as well as less than zero and then you remove the redundancies. E.g.: x>2 and x>3 implies only x>3 as seen here: https://www.khanacademy.org/math/algebra-home/alg-quadratics/alg-quadratic-inequalities/v/quadratic-inequality-example at the 5 minute mark

Since when do inequalities assume something is one-to-many?

Also, \(x\) is certainly not two numbers in a one-to-many graph. One-to-many just means that for each \(x\) value there's more than one \(y\) value associated to it, implying that the relation isn't a function to begin with.

What you've learnt from Khanacademy is the exact same thing TheAspiringDoc tried to say.

[Mattjbr2]

Since when do inequalities assume something is one-to-many?

Also, \(x\) is certainly not two numbers in a one-to-many graph. One-to-many just means that for each \(x\) value there's more than one \(y\) value associated to it, implying that the relation isn't a function to begin with.

What you've learnt from Khanacademy is the exact same thing TheAspiringDoc tried to say.

What? I didn't assume that. My questions and examples aren't being inferred the way I implied them. Never mind! I've figured out the issue. Thanks anyway

For anyone interested, this is how you solve this in the future. This is what I should have been told today.

MOD EDIT: double posting

[darkz]

For anyone interested, this is how you solve this in the future. This is what I should have been told today.

Or you can just do it with the graphical method by inspection since that way is much quicker in my opinion

[DBA-144]

What? I didn't assume that. My questions and examples aren't being inferred the way I implied them. Never mind! I've figured out the issue. Thanks anyway

For anyone interested, this is how you solve this in the future. This is what I should have been told today.

MOD EDIT: double posting

Isnt x is a set of r not including the interval a to b inclusive the same as x is greater than a and x is less than b

Sorry dont know how to do the actual math thing

[Mattjbr2]

Isnt x is a set of r not including the interval a to b inclusive the same as x is greater than a and x is less than b

Sorry dont know how to do the actual math thing

Yup; either would suffice

[Hala119]

Hey guys, can anyone help with this problem?

Find the length cut off the x-axis and y-axis by the circle x^2 + y^2 - 2x - 4y = 20?

[jazzycab]

Hey guys, can anyone help with this problem?

Find the length cut off the x-axis and y-axis by the circle x^2 + y^2 - 2x - 4y = 20?

This seems a strangely worded question, but it sounds like you're looking for the distance between the \(x\)-intercepts and the distance between the \(y\)-intercepts:

So the distance between \(x\)-intercepts is \(1+\sqrt{21}-\left(1-\sqrt{21}\right)=2\sqrt{21}\).
You can do similar for the distance between \(y\)-intercepts.