Solving for X in a 60x^2 - 90x^4 Equation

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To solve the equation 0 = 60x^2 - 90x^4, it's essential to factor it rather than divide by terms that could be zero. The equation can be rewritten as 0 = (2 - 3x^2)30x^2, indicating that either 30x^2 = 0 or 2 - 3x^2 = 0 must hold true. This leads to the solutions x = 0 or x^2 = 2/3, giving x = ±√(2/3). Dividing by zero is avoided by recognizing that if 30x^2 = 0, x must be zero, which is a valid solution. The discussion emphasizes the importance of factoring and understanding the implications of multiplying by zero in equations.
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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 can't do that with a negative number. So.. yeah, i need help. my algebra skills are very rusty...thanks
 
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Maybe writing out your steps would help...


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

90x^4 - 60^2
1.5x^4 = x^2
1.5 = 1/(x^2)
1.5 = -x^2
the negative root?
 
I don't like your last step. -x^2 = 1 / x^2 is never right.
 
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.
 
robert Ihnot as long as x is not = 0


try this:
0 = 60x^2 - 90x^4
0 = (2 - 3x^2)30x^2
 
Last edited:
JonF: robert Ihnot as long as x is not = 0. OF COURSE! And if x IS ZERO, well then that's an answer too!
 
in which step do i divide by 30x^2? right when i divide by 60x^2?
 
i knew there was factoring involved, i just couldn't get it to factor right, but then i still have to divide both sides, and dividing zero?
 
  • #10
ahh perfect, dividing by 30x works just fine, but i still have the negative there?
 
  • #11
Then you want to solve for that factor. Look again at JonF.
 
  • #12
ok...gracias
 
  • #13
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 isn’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.
 

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