VCE Methods Question Thread!

[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?

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.

There is only one type of square root of a positive real number. It is denoted as V2, and it is always positive.
Any positive real number has two real square roots, they are V2 and -V2.

[paulsterio]

OR

The square root of a number is always positive. This is why there is a "square-root function" - which is one-to-one. If we were to say what numbers squared to give 1 for instance, which is solved above, the answer will be + square root of 1 OR - square root of 1. The square root of 1 is always positive. If that makes much sense.

[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

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.

There is only one type of square root of a positive real number. It is denoted as V2, and it is always positive.
Any positive real number has two real square roots, they are V2 and -V2.

ok when i said 2 types, i meant there are 2 definitions for a "square root", the principle square root is commonly called "the" square root.

trivial

[yawho]

What are the two definitions for a 'square root'?

[TrueTears]

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

y bother over something so trivial though?

[paulsterio]

What are the two definitions for a 'square root'?

It really doesn't matter

[Insa]

Hey guys,

I need some help with this question. What do I do?

A function g(x) is mapped to the curve h(x)= -g(4(x+1))+3. Create a matrix equation that will map g(x) to h(x).

Thanks 

[Phy124]

Hey guys,

I need some help with this question. What do I do?

A function g(x) is mapped to the curve h(x)= -g(4(x+1))+3. Create a matrix equation that will map g(x) to h(x).

Thanks 

Ok so I transformation matrix looks like this:



Where;
- a produces a dilation by a factor of "a" in the x-direction (*Note: a negative if front of the "a" will produce a reflection about the y-axis)
- b produced a dilation by a factor of "b" in the y-direction (*Note: a negative if front of the "b" will produce a reflection about the x-axis)
- c translates the graph "c" units in the x-direction
- d translates the graph "d" units in the y-direction

We have:
- a dilation by a factor of 1/4 in the x-direction
- a reflection about the x-axis
- a translation 1 unit to the left
- a translation 3 units up

So the transformation matrix will be:



Just in case you don't understand, I'll show you why this works:

Let y = g(x)

Solve the matrix for x' and y'





Rearrange to have x in terms of x' and y in terms of y'





Sub x and y back into original equation

[Insa]

Cheers Phy124! Appreciate your help

[yawho]

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

y bother over something so trivial though?

It was trivial until you said 'there are two definitions for square root of a positive number'.

[TrueTears]

ok.

[ashoni]

i need help with this question.
Find linear factors:
5(x-3)^2 -15

show working out if possible thanks

[noname]

Could someone help me with part a and part b of:


Much appreciated

[brightsky]

i need help with this question.
Find linear factors:
5(x-3)^2 -15

show working out if possible thanks

5((x-3)^2 - 3) = 5(x-3+sqrt(3))(x-3-sqrt(3))

Could someone help me with part a and part b of:
(Image removed from quote.)

Much appreciated

12. a) f(x)=f(-x)
so (px+q)/(x+r)=(-px+q)/(-x+r) for all x
(px+q)(-x+r)=(-px+q)(x+r)
-px^2 + prx -qx + qr = -px^2 -prx + qx + qr
prx - qx = -prx + qx
pr - q = -pr + q
2pr = 2q
pr = q
sub this into f(x)
f(x) = (px + pr)/(x+r) = p(x+r)/(x+r) = p as required

b) f(-x)=f(-x)
(-px + q)/(-x+r)=(-px-q)/(x+r)
(-px+q)(x+r)=(-px-q)(-x+r)
-px^2 - prx + qx + qr = px^2 - prx + qx - qr
-px^2 + qr = px^2 - qr
2px^2 = 2qr
px^2 = qr
p = qr/x^2
sub this into f(x)
f(x)=[(qr/x^2)x + q]/(x+r)
= (qr/x + q)/(x+r)
= (qr + qx)/[x(x+r)]
= [q(r+x)]/[x(x+r)]
= q/x

[noname]

i need help with this question.
Find linear factors:
5(x-3)^2 -15

show working out if possible thanks

5((x-3)^2 - 3) = 5(x-3+sqrt(3))(x-3-sqrt(3))

Could someone help me with part a and part b of:
(Image removed from quote.)

Much appreciated

12. a) f(x)=f(-x)
so (px+q)/(x+r)=(-px+q)/(-x+r) for all x
(px+q)(-x+r)=(-px+q)(x+r)
-px^2 + prx -qx + qr = -px^2 -prx + qx + qr
prx - qx = -prx + qx
pr - q = -pr + q
2pr = 2q
pr = q
sub this into f(x)
f(x) = (px + pr)/(x+r) = p(x+r)/(x+r) = p as required

b) f(-x)=f(-x)
(-px + q)/(-x+r)=(-px-q)/(x+r)
(-px+q)(x+r)=(-px-q)(-x+r)
-px^2 - prx + qx + qr = px^2 - prx + qx - qr
-px^2 + qr = px^2 - qr
2px^2 = 2qr
px^2 = qr
p = qr/x^2
sub this into f(x)
f(x)=[(qr/x^2)x + q]/(x+r)
= (qr/x + q)/(x+r)
= (qr + qx)/[x(x+r)]
= [q(r+x)]/[x(x+r)]
= q/x

You're amazing, thanks for the help