Pandalover5165 Pandalover5165
  • 22-10-2019
  • Mathematics
contestada

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

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nuuk nuuk
  • 22-10-2019

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]

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