VCE Methods Question Thread!

[Eiffel]

so we know p(2m) = -2
a(4m^2) +6m - 4 = -2
4am^2 +6m - 2 = 0

is it right to now treat m as our x? so

b^2 - 4ac = 0
36 - 4(4a)(-2) = 0
36 + 32a = 0
32a = -36
a = -9/8 ??

[knightrider]

For the equation when finding the stationary points and determining their nature.

Would you say

SP at (0,8) is a local maximum

or would you say SP at (0,8) is a maximum

because in this case the SP will be the maximum so do we say local maximum or  maximum?

[keltingmeith]

For the equation when finding the stationary points and determining their nature.

Would you say

SP at (0,8) is a local maximum

or would you say SP at (0,8) is a maximum

because in this case the SP will be the maximum so do we say local maximum or  maximum?

Both are correct.

[StressedAlready]

Solve the simultaneous equations y = mx and y = (1/x) + 5 for x in terms of m.

I've tried it but... in the answers they give two sets of y = _____________ and x= __________ answers.Why do they do this? Aren't they only looking for x = _______________ in terms of m?

[keltingmeith]

Just listing x=____ would be fine.

[cosine]

Find a such that f(x) has two solutions

This is the exact question my school gave us. Is this a stupid question? I mean, it doesn't really tell you what it means and what it wants.. for all i did

(ax  +   5)(x    -  1)

So, if a=1 there can be two solutions..

But apparently we have to use the discriminant..

[keltingmeith]

Find a such that f(x) has two solutions

This is the exact question my school gave us. Is this a stupid question? I mean, it doesn't really tell you what it means and what it wants.. for all i did

(ax  +   5)(x    -  1)

So, if a=1 there can be two solutions..

But apparently we have to use the discriminant..

I reckon it's incredibly valid - the question's a little vague, sure, but I wouldn't give your solution full marks simply because you've factorised wrong... Try expanding, see what comes out. Discriminant is definitely the way to go.

[StupidProdigy]

Find a such that f(x) has two solutions

I reckon it's incredibly valid - the question's a little vague, sure, but I wouldn't give your solution full marks simply because you've factorised wrong... Try expanding, see what comes out. Discriminant is definitely the way to go.

Being curious about methods and whatnot, I decided to try this question because math quest doesn't have much on it, I got a>-4/5
Is that right? just curious hahaha, thanks

[cosine]

Being curious about methods and whatnot, I decided to try this question because math quest doesn't have much on it, I got a>-4/5
Is that right? just curious hahaha, thanks

correct, how though?

[keltingmeith]

correct, how though?

The discriminant, b^2-4ac.

If b^2-4ac>0, we have two solutions, so:

4^2-4(a)(-5)>0
16+20a>0
20a>-16
a>-16/20
a>-4/5

[StupidProdigy]

correct, how though?

Well using the discriminant we know that b^2 -4ac must be great than zero for two solutions right
So with you're equation, a=a, b=4, c=-5
Therefore (4)^2 -4(a)(-5)>0
                 16+20a>0
                 20a>-16
                 a>-16/20
                 a>-4/5

edit: beaten to it by Euler haha!

[StressedAlready]

Just listing x=____ would be fine.

Then why do the answers list two x = ______________ answers and two y = _________________ answers?

[keltingmeith]

Then why do the answers list two x = ______________ answers and two y = _________________ answers?

Because the simultaneous equations have two solutions. The reason they've listed the y=_________ is beyond me (given how they're framed the question), but the reason there's two sets is because there's two solutions.

[cosine]

Why you guys use the discriminant formula?

[keltingmeith]

Why you guys use the discriminant formula?

We wanted to find what values of a gave two solutions.
The discriminant can be used to find when the general polynomial has 0, 1 or 2 solutions. So, we used it.