VCE Methods Question Thread!

[cnguyen599]

Thanks a bunch! It seems so simple now.

[Kanon]

this has me so frustrated haha

Find the values of p for which has 2 real solutions

I'm betting this requires you to use the quadratic solution as it's asking for two real solutions.  To use the quadratic solution you must have a quadratic, which 2x^2 is.
However the p's in the brackets have me confused, what i've been trying to do is get the p's in singular terms as to say that 1-2p = 0 therefore, p = 1/2.  I'm not sure if i'm on the right track, could someone please tell me if that's the most logical way of approaching the problem?  The only other thing i can think of is using Sim Eqn.

[Milkshake]

this has me so frustrated haha

Find the values of p for which has 2 real solutions

I'm betting this requires you to use the quadratic solution as it's asking for two real solutions.  To use the quadratic solution you must have a quadratic, which 2x^2 is.
However the p's in the brackets have me confused, what i've been trying to do is get the p's in singular terms as to say that 1-2p = 0 therefore, p = 1/2.  I'm not sure if i'm on the right track, could someone please tell me if that's the most logical way of approaching the problem?  The only other thing i can think of is using Sim Eqn.

Using the discriminant formula should get you the answer, as it tells us how many real solutions(x-intercepts) the quadratic has depending on the value it gives. Discriminant: b² - 4ac > 0 (since the question states we must have 2 real solutions)
Then you just plug in the a, b and c values and you're good to go! a=2(1-2p) etc etc

[Genericname2365]

Really simple question here (year 11 stuff really ), but I just need to be refreshed on the order in which I need to solve this.

f(a+2)=2(a+2)²-6(a+2)+1

I ended up with 2a²-6a-3
but it should be 2a²+2a-3. How do I get that?

[b^3]

Firstly note that this is simplfying, not solving (solving would require us to find an expression for a in this situation)


EDIT: Looking at it you probably missed the 4a in the expansion of

[Genericname2365]

Firstly note that this is simplfying, not solving (solving would require us to find an expression for a in this situation)


EDIT: Looking at it you probably missed the 4a in the expansion of

Thanks, it makes sense now.

[#1procrastinator]

Why is the square root of a number always positive and not plus or minus?

[TrueTears]

just a misunderstanding of definitions, there are 2 types of square roots, one is called the principle square root (which we always call it just the 'square root') and the principle square root is always defined to be the positive.

but in general if we are talking about the 'general square root', then it needs to be positive and negative.

http://en.wikipedia.org/wiki/Square_root

[#1procrastinator]

How come we're told only to put down the principle square root in methods? (at least in my class)

[Kanon]

Just a few general questions

Is there a more time-efficient manner of approaching this question?

Find the linear factors of:   

I know when given a polynomial, it's best to look at the factors of the last term (the one without an x) and sub them into the Polynomial Function.
I was doing this without a calculator, and I got to about 4 and -4 when I was just like cbf, i'll look for a linear term at the back of the book and just use that. 
What's the likelihood of something like this popping up as a no-tech exam question?

[brightsky]

-4x^3 - 30x^2 - 36x = -2x(2x^2 +15x + 18) = -2x(2x^2+15x+18)
now you need only to deal with the quadratic 2x^2 +15x + 18, which you can just factorise.

[Kanon]

-4x^3 - 30x^2 - 36x = -2x(2x^2 +15x + 18) = -2x(2x^2+15x+18)
now you need only to deal with the quadratic 2x^2 +15x + 18, which you can just factorise.

Just to reaffirm, we're only able to do this because their is no x-term, and as such we can take a common factor of 2x out of the equation?

If so, I shall name my firstborn after you.

[brightsky]

what do you mean 'no x-term'? and yeah always factorise if you can. but alas, for general polynomials, more often than not you will be forced into crude trial and error techniques exploiting the factor theorem.

[yawho]

just a misunderstanding of definitions, there are 2 types of square roots, one is called the principle square root (which we always call it just the 'square root') and the principle square root is always defined to be the positive.

but in general if we are talking about the 'general square root', then it needs to be positive and negative.

http://en.wikipedia.org/wiki/Square_root

there are 2 types of square roots, one is called the principle square root , what is the other type?
What are the notations for the two types?

[brightsky]

just a misunderstanding of definitions, there are 2 types of square roots, one is called the principle square root (which we always call it just the 'square root') and the principle square root is always defined to be the positive.

but in general if we are talking about the 'general square root', then it needs to be positive and negative.

http://en.wikipedia.org/wiki/Square_root

there are 2 types of square roots, one is called the principle square root , what is the other type?
What are the notations for the two types?

negative square root i guess? i think you're over complicating things. when we 'take the square root' of a number, we just put a square root over that number. we never randomly add a negative sign in front. this is just like 'squaring' a number or taking the mod of a number, we don't just randomly add a negative sign.