Quote by autodidude
[k+1±√((k^2+130k)/4k^2)]/4k

I'm not really sure how you got that... It's not right first of all, but then you get the right answer later down the track?
From [tex]2kx^2(k+1)x+4=0[/tex] we have
[tex]a=2k[/tex]
[tex]b=(k+1)[/tex]
[tex]c=4[/tex]
Now, plugging into the quadratic [tex]x=\frac{b\pm\sqrt{b^24ac}}{2a}[/tex]
we get [tex]\frac{k+1\pm\sqrt{(k+1)^24(2k)(4)}}{2(2k)}[/tex]
Simplifying, we get [tex]x=\frac{k+1\pm\sqrt{k^230k+1}}{4k}[/tex]
If you try to use the quadratic formula by first dividing through by the coefficient of x
^{2}, then
[tex]a=1[/tex]
[tex]b=\frac{(k+1)}{2k}[/tex]
[tex]c=\frac{2}{k}[/tex]
[tex]x=\frac{\frac{k+1}{2k}\pm\sqrt{\left(\frac{k+1}{2k}\right)^2\frac{8}{k}}}{2}[/tex]
So now you should be able to take out a factor of [tex]\frac{1}{2k}[/tex] from the surd, and bring that down to the denominator of the fraction to get the same result as before.