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

  • Context: High School 
  • Thread starter Thread starter ziddy83
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Discussion Overview

The discussion revolves around solving the equation 0 = 60x^2 - 90x^4. Participants explore various methods for finding the values of x, including factoring and addressing issues related to dividing by zero.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses difficulty in solving the equation and mentions rusty algebra skills.
  • Another suggests writing out steps and advises against dividing by an indeterminate quantity.
  • A participant attempts to manipulate the equation but makes an error in their reasoning regarding negative roots.
  • Concerns are raised about the validity of dividing by 30x^2, with a participant emphasizing that this should not be done unless x is not equal to zero.
  • Some participants propose factoring the equation as a more effective approach to finding solutions.
  • There is a discussion about the implications of multiplying two quantities to equal zero, leading to potential solutions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method for solving the equation, and there are disagreements regarding the validity of certain steps, particularly around dividing by zero and the interpretation of negative roots.

Contextual Notes

There are unresolved issues regarding the assumptions made when dividing by terms that could be zero, as well as the implications of negative values in the context of the equation.

ziddy83
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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 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. [itex]-x^2 = 1 / x^2[/itex] 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:
[tex]0 = 60x^2 - 90x^4[/tex]
[tex]0 = (2 - 3x^2)30x^2[/tex]
 
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: [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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