# Math Help - Help Needed

1. ## Help Needed

First on a fuction f(x)=2x^2-3
How do I find x such that f(x)=x

How do I factorise 3x^2y-6xy+12xy^2

And solve the equation
1-3y=2(y-8)

2. Originally Posted by t=0m
First on a fuction f(x)=2x^2-3
How do I find x such that f(x)=x
You are looking for a value of x such that
$2x^2 - 3 = x$

or
$2x^2 - x - 3 = 0$

See what you can come up with. (Hint: the quadratic factors.)

-Dan

3. Originally Posted by t=0m
How do I factorise 3x^2y-6xy+12xy^2
All three terms contain a common 3xy. So:
$3x^2y-6xy+12xy^2 = 3xy(x - 2 + 4y)$
is the best we can do.

-Dan

4. Originally Posted by t=0m
And solve the equation
1-3y=2(y-8)
Come now. If you are supposed to be able to do the other two you should be able to do this one.

I'll start you off:
$1-3y=2(y-8)$

$1 - 3y = 2y - 16$

Can you take it from here?

-Dan

5. The funtion is the top one and you need to make The number you put in (x) = x

6. Originally Posted by topsquark
You are looking for a value of x such that
$2x^2 - 3 = x$

or
$2x^2 - x - 3 = 0$

See what you can come up with. (Hint: the quadratic factors.)

-Dan
Originally Posted by t=0m
The funtion is the top one and you need to make The number you put in (x) = x
Isn't that what I did?

-Dan