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Solving for x

  1. Dec 7, 2004 #1
    hi, I was wonderin how i would solve for X in the following equation...

    [tex] 0 = 60x^2 - 90x^4 [/tex]

    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. jcsd
  3. Dec 7, 2004 #2


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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.
  4. Dec 7, 2004 #3
    as you wish..............................

    90x^4 - 60^2
    1.5x^4 = x^2
    1.5 = 1/(x^2)
    1.5 = -x^2
    the negative root?
  5. Dec 7, 2004 #4


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    I don't like your last step. [itex]-x^2 = 1 / x^2[/itex] is never right.
  6. Dec 7, 2004 #5
    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.
  7. Dec 7, 2004 #6
    robert Ihnot as long as x is not = 0

    try this:
    [tex] 0 = 60x^2 - 90x^4 [/tex]
    [tex] 0 = (2 - 3x^2)30x^2 [/tex]
    Last edited: Dec 7, 2004
  8. Dec 7, 2004 #7
    JonF: robert Ihnot as long as x is not = 0. OF COURSE!!! And if x IS ZERO, well then that's an answer too!
  9. Dec 7, 2004 #8
    in which step do i divide by 30x^2? right when i divide by 60x^2?
  10. Dec 7, 2004 #9
    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?
  11. Dec 7, 2004 #10
    ahh perfect, dividing by 30x works just fine, but i still have the negative there?
  12. Dec 7, 2004 #11
    Then you want to solve for that factor. Look again at JonF.
  13. Dec 7, 2004 #12
  14. Dec 8, 2004 #13
    You should never divide by 30x^2 as hurkyl suggested.

    If you are with me up to: [tex] 0 = (2 - 3x^2)30x^2 [/tex]...

    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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