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Mentallic
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#9
Feb24-11, 12:40 AM
HW Helper
P: 3,436
Quote Quote by autodidude View Post
Well I did get this: [tex]x=\frac{\frac{k+1}{2k}\pm\sqrt{\left(\frac{k+1}{2k}\right)^2-\frac{8}{k}}}{2}[/tex]


but then I expanded ((k+1)/2k)^2 to get (k+1)^2/4k^2 and multiplied 8/k by 4k/4k to get 32k/4k^2





Could you elaborate please? (baby steps lol)
Yep that's right, so what you have so far is

[tex]
x=\frac{\frac{k+1}{2k}\pm\sqrt{\frac{\left(k+1\right)^2}{4k^2}-\frac{32k}{4k^2}}}{2}
[/tex]

Looking at the discriminant only:

[tex]\Delta=\frac{\left(k+1\right)^2-32k}{4k^2}[/tex]

[tex]=\frac{k^2+2k+1-32k}{4k^2}[/tex]

[tex]=\frac{k^2-30k+1}{4k^2}[/tex]

Now what I mean by taking out a factor of [tex]\frac{1}{2k}[/tex] is that [tex]\sqrt{4k^2}=2k[/tex] so we have:

[tex]
x=\frac{\frac{k+1}{2k}\pm\frac{\sqrt{k^2-30k+1}}{\sqrt{4k^2}}}{2}
[/tex]

[tex]
x=\frac{\frac{k+1}{2k}\pm\frac{\sqrt{k^2-30k+1}}{2k}}{2}
[/tex]

Now we can factor out [tex]\frac{1}{2k}[/tex] from the numerator, and we can move this to the denominator to get our desired result.