VCE Methods Question Thread!

[shadows]

Random question.


Will I lose an answer mark if I leave it not simplified..


72pi/4? (This is exam 1 btw)

[Sanguinne]

This question is from a MAV exam 1

For what values of k, where k is a real constant, does the equation

i let to make a quadratic and hence use the discriminant. However the answer I found for k was when it is actually

Why is this so?

[clıppy]

This question is from a MAV exam 1

For what values of k, where k is a real constant, does the equation

i let to make a quadratic and hence use the discriminant. However the answer I found for k was when it is actually

Why is this so?

b^3 already helped me with that

Re: Methods [3/4] Question Thread!

[Alwin]

Random question.

Will I lose an answer mark if I leave it not simplified..

72pi/4? (This is exam 1 btw)

Personally I would simplify. I think that one time my teacher did take marks of for people that didn't cancel a fraction like that in a sac, but dunno if they would deduct a mark (or 1/2 a mark which is basically the same) on the exam

This question is from a MAV exam 1

For what values of k, where k is a real constant, does the equation

i let to make a quadratic and hence use the discriminant. However the answer I found for k was when it is actually

Why is this so?

Does the equations: .......... ?? I'm guessing from memory that the question finishes with "2 solutions" ?

Hint: remember that So then both solutions have to be positive


@Ahat: nice vocag but a tad busy atm so gimme a bit of time and I'll post a better solution that isn't all over the place like my first one was. sorry mate!

[darklight]

A man 2m tall walks away from a lamppost whose light is 5m above the ground. If he walks at a speed of 1.5m/s, at what rate is his shadow growing when he is 10m from the lamppost?

[b^3]

We basically have the situation below. Press the a slider
https://www.desmos.com/calculator/u6wwabczmk
The blue line is the man and the red line is the lamppost.

EDIT1: Well that went wrong quickly...

EDIT2: Used the wrong height of the lamppost, fixed in the diagram now. Do what RKTR said.

[RKTR]

use similar triangles

Let length of shadow = x , distance from lamp post = y

x/2 =(x+y)/5

5x=2x+2y
3x=2y
x=2/3 y

[ahat]

If I had the quadratic function:

First, let:
u = log_e(5)
v = log_e(x)

0 = 2u² - 5uv + 2v²
And then I used the quadratic formula:

Let a = 2, b = -5, c = 2

so

Quadratic formula:

 
=
=

So the solutions are

or

My question, what do the solutions represent? u or v or x?

Thanks

[Phy124]

If I had the quadratic function:

I think you're missing some key information here?

Edit: Oh, I think I see what you've done

You have ?

If you're going to use u's and v's I would just go straight from to So you know that and and hence

[Stevensmay]

I'm getting from the equation you gave later on.

[b^3]

, that is is a constant. So when you use the quadratic formula here you need to look at as your variable and as a constant that can be contained in the coefficients.

Then solve for if you need to, depending on what the question is.
You could also do it by completing the square.


I'm not entirely sure what your question is asking, but that's how'd you would apply the quadratic formula given what information we have.

[Alwin]

to shorten the quote

If I had the quadratic function:

First, let:
u = log_e(5)
v = log_e(x)

0 = 2u² - 5uv + 2v²
And then I used the quadratic formula:

Let a = 2, b = -5, c = 2

so

Quadratic formula:

 
=
=

So the solutions are

or

My question, what do the solutions represent? u or v or x?

Thanks

Hmm, not too thrilled about your use of the quadratic formula.

I would have done:




Subbing in u = log_e(5) and v = log_e(x),





EDIT: beaten same answers as Stevensmay got

[ahat]

Hahaha, thanks guys. The full question:

Solve the equation for x.

Like Phy124 said, I got this expression

Edit: Oh, I think I see what you've done
You have ?

and then I used the quadratic formula because that's the only way I could think of for getting x, but I didn't know what the 2 or 0.5 represented.

, that is is a constant. So when you use the quadratic formula here you need to look at as your variable and as a constant that can be contained in the coefficients.
[tex]\begin{alignedat}{1}2v^{2}+5uv+2u^{2} & =0
\\ a=2,\: b=5u,\: c=2u^{2}

Thanks, makes sense

[Alwin]

Hahaha, thanks guys. The full question:

Solve the equation for x.

Like Phy124 said, I got this expression
and then I used the quadratic formula because that's the only way I could think of for getting x, but I didn't know what the 2 or 0.5 represented.

You could have gone:


Then solve the quadratic and re-substitute back in and make sure your x values are positive

[ahat]

@Ahat: nice vocag but a tad busy atm so gimme a bit of time and I'll post a better solution that isn't all over the place like my first one was. sorry mate!

Btw, thanks Alwin, just read this. If you do get a chance to help, you can ignore the general solution I posted, I know I did it wrong
(to look at it optimistically, at least I'm not going to lose marks on general solutions now, knowing my mistake )

You could have gone:


Then solve the quadratic and re-substitute back in and make sure your x values are positive

Holy moly, sweet! I always thought that the base had to be e if you're changing to that form, ahh, that makes it so much easier. Much appreciated man ♥

Mod Edit: Merged double post (Phy124)

EDIT SPREE: What's the range of possibilities the base can be changed to?