Quadratics with fractions
- Thread starter spear33
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You've followed the tutor's instructions (to multiply through by k<sup>2</sup> to clear out the denominators); now rearrange the quadratic in the usual form, and apply the usual methods for solving. The Quadratic Formula, for instance, will work, as will factoring.
Eliz.
Eliz.
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How are you getting that (2)(k<sup>2</sup>) equals 2, or that (-5/k)(k<sup>2</sup> is still -5/k...? Did you forget that "multiplying through on both sides" means "multiplying through onto every term on each side"?spear33 said:
Note: (0)(k<sup>2</sup>) = 0.
Eliz.
Edit: Correcting formatting.
To do it literally, multiplying both sides by \(\displaystyle \mbox{k^2}\) is to go:
\(\displaystyle \mbox{ k^2\left(2 - \frac{5}{k} + \frac{2}{k^2}\right) = k^2(0) }\)
where \(\displaystyle \mbox{k^2(0) = 0}\) and \(\displaystyle \mbox{ k^2\left(2 - \frac{5}{k} + \frac{2}{k^2}\right)}\) expands to?
\(\displaystyle \mbox{ k^2\left(2 - \frac{5}{k} + \frac{2}{k^2}\right) = k^2(0) }\)
where \(\displaystyle \mbox{k^2(0) = 0}\) and \(\displaystyle \mbox{ k^2\left(2 - \frac{5}{k} + \frac{2}{k^2}\right)}\) expands to?
Yeah, actually -5 was written as a positive 5 on my paper, but in my rush to post it I thought I accidently forgot to put the - in front on my paper. Anyways thanks for all the help and I'm sure I'll be back here sometime this semester (probably tonight).
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No prob; no worries. But if you would like to learn how to format with LaTeX (and if I can do it, it can't be that hard!), then follow the lower links in the "Forum Help" pull-down menu at the very top of the page.spear33 said:
Eliz.