Question

For what value of k are there two complex solutions to the given quadratic equation (k+1)x²+4kx+2=0.

Answer 1

Answer:$k\in \left ( \frac{-1}{2},1\right )$

Step-by-step explanation:

Given

$\left ( k+1\right )x^2+4kx+2=0$

For complex  roots Discriminant should be zero

D<0

$D=\sqrt{b^2-4ac}$

here

$D=\sqrt{\left ( 4k\right )^2-4\left ( k+1\right )\left ( 2\right )}$

$D=\sqrt{16k^2-8k-8}$

so $16k^2-8k-8$

$2k^2-k-1$

$\left ( 2k+1\right )\left ( k-1\right )$

so $k\in \left ( \frac{-1}{2},1\right )$