Asked by Kaur
                for what values of k roots of
2x^2 + 4k x + 128 = 0 are imaginary
            
        2x^2 + 4k x + 128 = 0 are imaginary
Answers
                    Answered by
            oobleck
            
    c'mon, man. I know that yo are aware of the discriminant for quadratics. Complex roots occur when it is negative. Thatis, when
(4k)^2 - 4*2*128 < 0
k^2 < 64
|k| < 8
confirm this by trying a couple of values.
    
(4k)^2 - 4*2*128 < 0
k^2 < 64
|k| < 8
confirm this by trying a couple of values.
                    Answered by
            Anonymous
            
    by the way first divide the whole mess by 2 to simplify your life.
    
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