Answer:[tex]k\in \left ( \frac{-1}{2},1\right )[/tex]
Step-by-step explanation:
Given
[tex]\left ( k+1\right )x^2+4kx+2=0[/tex]
For complex roots Discriminant should be zero
D<0
[tex]D=\sqrt{b^2-4ac}[/tex]
here
[tex]D=\sqrt{\left ( 4k\right )^2-4\left ( k+1\right )\left ( 2\right )}[/tex]
[tex]D=\sqrt{16k^2-8k-8}<0[/tex]
so [tex]16k^2-8k-8<0[/tex]
[tex]2k^2-k-1<0[/tex]
[tex]\left ( 2k+1\right )\left ( k-1\right )<0[/tex]
so [tex]k\in \left ( \frac{-1}{2},1\right )[/tex]
