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HW Helper
P: 3,531
 Quote by autodidude [k+1±√((k^2+1-30k)/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 $$2kx^2-(k+1)x+4=0$$ we have
$$a=2k$$
$$b=-(k+1)$$
$$c=4$$

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

we get $$\frac{k+1\pm\sqrt{(k+1)^2-4(2k)(4)}}{2(2k)}$$

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

If you try to use the quadratic formula by first dividing through by the coefficient of x2, then
$$a=1$$
$$b=\frac{-(k+1)}{2k}$$
$$c=\frac{2}{k}$$

$$x=\frac{\frac{k+1}{2k}\pm\sqrt{\left(\frac{k+1}{2k}\right)^2-\frac{8}{k}}}{2}$$

So now you should be able to take out a factor of $$\frac{1}{2k}$$ from the surd, and bring that down to the denominator of the fraction to get the same result as before.