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Feb23-11, 08:14 PM
HW Helper
P: 3,531
Quote Quote by autodidude View Post
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

Now, plugging into the quadratic [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

we get [tex]\frac{k+1\pm\sqrt{(k+1)^2-4(2k)(4)}}{2(2k)}[/tex]

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

If you try to use the quadratic formula by first dividing through by the coefficient of x2, then


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.