Re: Simple algebra question
- From: Chris Smith <cdsmith@xxxxxxx>
- Date: Tue, 21 Aug 2007 17:09:04 -0600
Adam K. <akress@xxxxxxxxx> wrote:
(x4 is x to the 4th, not x times 4). I'm not sure what is supposed to
happen to the first x in the equation when you divide the x² and
everything else by 4.
It's just x^4 / 4, which can't be simplified.
There are other responses there (Arturo's, for example, will work great
if you can follow it). Here's another way of solving the problem, which
is how I did it when I saw your post. The polynomial x^4 - 4x^2 + 3 can
be factored. Specifically,
x^4 - 4x^2 + 3 = (ax^2 + b)(cx^2 + d)
for some a, b, c, and d which should be fairly easy to find; just play
around a bit. Now the product of (ax^2 + b) and (cx^2 + d) will be zero
*precisely* when one of those two smaller factors are equal to zero. So
now you've got two quadratic equations to solve instead of one of those
ugly quartic (x^4) equations. Quadratic equations should be easy,
relatively speaking; remember that if a term is missing (b), it's
coefficient is zero.
--
Chris Smith
.
- References:
- Simple algebra question
- From: Adam K.
- Simple algebra question
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