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Mentallic
#7
Feb23-11, 08:14 PM
HW Helper
P: 3,531
Quote Quote by autodidude View Post
[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 [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^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
[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.