help!

Question: By completing the square, find in terms of the constant k the roots of the equation : x² + 4kx - k = 0
Hence find the set of values of k for which the equation has no real roots

i can't seem to do this and ive been trying for ages! Please help! Any help will do

x² +4kx- k =(x+2k)² -k-(2k)²
(x+2k)² = k+ 4k²
x+2k= ± √(k+4k² )
x= -2k ± √(k+4k² )

thanks a lot everyone! I think i understand it now

I'm not quite sure how to do it by completing the square, but there is a handy tool called the 'discriminant' that you can employ when solving problems like these.

x² + 4kx - k = 0

D = b² - 4ac

D = 16k² -4*1*(-k) = 16k² + 4k

For the equation to have no real roots, D < 0.

So, 16k² + 4k < 0

4k(4k + 1) < 0

-0.25 < k < 0

So, when -0.25 < k < 0, the equation has no real roots.

Hope this helps,

~~Simba

...

Wrong, 0<k<0.25

this is a 2004 thread !!