# Find the zeros of the function algebraically

1. Jan 21, 2010

### PanTh3R

1. The problem statement, all variables and given/known data
Find the zeros of the function algebraically

2. Relevant equations
f(x)=x/(9x^2-4)
and
f(x)=x^3-x

3. The attempt at a solution
i was gone for about a month and a half from the class because i had to move and im kinda rusty
ATTEMPT-

f(x)=x/(9x^2-4)
Multiplied by x
subtracted 9x^2
and added 4, which gave me.....
-9x^2+x+4=f(x) then i put it in a quadratic formula to find it and i got the wrong answer (i checked in the back of the book for the answer which was 0, i know im probly really wrong but i was away from the class a long time lol.

Second problem
f(x)=x^3-x
I honestly completely forgot how to solve for the zeros of a cubic function sorry :(

2. Jan 21, 2010

### Staff: Mentor

For the first one, the only time a rational expression (e.g., x/(9x^2 - 4)) can be zero is when the numerator is zero.

This doesn't make any sense.

3. Jan 21, 2010

### Staff: Mentor

For the 2nd problem, factor x^3 - x.

Frankly, missing a month and a half of a math class, regardless of whether the reason is good or not, might be insurmountable.

4. Jan 21, 2010

### PanTh3R

oops i put in the problem wrong and i figured it out(the second one)
f(x)=1/2x^3-x
x(1/2x^2-1)=0
x=0
1/2x^2-1=0
1/2x^2=1
x^2=2
x= +-square root of 2

5. Jan 21, 2010

### Char. Limit

Watch out on that second equation... you forgot about the x you factored out.

A cubic equation has three zeros.

6. Jan 22, 2010

### Mentallic

No he didn't!

it would suck if the OP were a girl. Still, I prefer assuming than typing in he/she.
They should really create a word to indicate either sex!