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 ( (-1)/(2),1\right )`

Explanation:

Given

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

For complex roots Discriminant should be zero

D<0

`D=√(b^2-4ac)`

here

`D=√(\left ( 4k\right )^2-4\left ( k+1\right )\left ( 2\right ))`

`D=√(16k^2-8k-8)<0`

so
`16k^2-8k-8<0`

`2k^2-k-1<0`

`\left ( 2k+1\right )\left ( k-1\right )<0`

so
`k\in \left ( (-1)/(2),1\right )`