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romanna [79]
4 years ago
7

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

Mathematics
1 answer:
nlexa [21]4 years ago
3 0

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 )

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