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 Quote by autodidude Well I did get this: $$x=\frac{\frac{k+1}{2k}\pm\sqrt{\left(\frac{k+1}{2k}\right)^2-\frac{8}{k}}}{2}$$ 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

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

Looking at the discriminant only:

$$\Delta=\frac{\left(k+1\right)^2-32k}{4k^2}$$

$$=\frac{k^2+2k+1-32k}{4k^2}$$

$$=\frac{k^2-30k+1}{4k^2}$$

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

$$x=\frac{\frac{k+1}{2k}\pm\frac{\sqrt{k^2-30k+1}}{\sqrt{4k^2}}}{2}$$

$$x=\frac{\frac{k+1}{2k}\pm\frac{\sqrt{k^2-30k+1}}{2k}}{2}$$

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