# Solving for x

1. Dec 7, 2004

### ziddy83

hi, I was wonderin how i would solve for X in the following equation...

$$0 = 60x^2 - 90x^4$$

If i do add 90x^4 from both sides....and then go about solving for x, i get 1/x, which is the same as -x...and then i have to take the square root of a number, and i cant do that with a negative number. So.. yeah, i need help. my algebra skills are very rusty...thanks

2. Dec 7, 2004

### Hurkyl

Staff Emeritus
Maybe writing out your steps would help...

P.S., instead of dividing by an indeterminate quantity, it's usually better to factor it out.

3. Dec 7, 2004

### ziddy83

as you wish..............................

90x^4 - 60^2
1.5x^4 = x^2
1.5 = 1/(x^2)
1.5 = -x^2
the negative root?

4. Dec 7, 2004

### Hurkyl

Staff Emeritus
I don't like your last step. $-x^2 = 1 / x^2$ is never right.

5. Dec 7, 2004

### robert Ihnot

As Ziddy83 is trying to pointing out AB = 0 implies A =0 or B=0. (Reams have been written on that kind of stuff) so if we divide by 30X^2, we arrive at a much simpler equation to work with.

6. Dec 7, 2004

### JonF

robert Ihnot as long as x is not = 0

try this:
$$0 = 60x^2 - 90x^4$$
$$0 = (2 - 3x^2)30x^2$$

Last edited: Dec 7, 2004
7. Dec 7, 2004

### robert Ihnot

JonF: robert Ihnot as long as x is not = 0. OF COURSE!!! And if x IS ZERO, well then that's an answer too!

8. Dec 7, 2004

### ziddy83

in which step do i divide by 30x^2? right when i divide by 60x^2?

9. Dec 7, 2004

### ziddy83

i knew there was factoring involved, i just couldnt get it to factor right, but then i still have to divide both sides, and dividing zero?

10. Dec 7, 2004

### ziddy83

ahh perfect, dividing by 30x works just fine, but i still have the negative there?

11. Dec 7, 2004

### robert Ihnot

Then you want to solve for that factor. Look again at JonF.

12. Dec 7, 2004

### ziddy83

ok...gracias

13. Dec 8, 2004

### JonF

You should never divide by 30x^2 as hurkyl suggested.

If you are with me up to: $$0 = (2 - 3x^2)30x^2$$...

When two quantities multiplied together equal zero that implies that one or both of them is zero.

So you have 30x^2 being multiplied to (2-3x^2).

This implies that if 30x^2 = 0 it would be a solution to your equation.

But also if 2-3x^2=0 that would also be a solution to your equation.

You can also think of it this way: You 30x^2 multiplied to some garbage, but it isnâ€™t going to mater what that garbage is if 30x^2 = 0, likewise you also have 2-3x^2 being multiplied to some junk. Who cares what that junk is if 2-3x^2=0, because than that junk is being multiplied by zero, and anything multiplied by zero is zero.