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jkasdhk

  • 2 years ago

minimum possible value of K for which the equation 2x^2+6x+K=0 has complex roots?

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  1. mukushla
    • 2 years ago
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    for having complex roots Discriminant of quadratic equation must be less than zero then what is result?

  2. jkasdhk
    • 2 years ago
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    its 9/2 (4.5)<0. Will the answer be 4?

  3. UnkleRhaukus
    • 2 years ago
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    \[2x^2+6x+K=0\] has complex roots when the discriminate is <0 \[\Delta=b^2-4ac\]\[=6^2-4\times1\times K\]

  4. UnkleRhaukus
    • 2 years ago
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    because in the quadratic formula we have to take the square root of the discriminant , \[x=\frac{-b\pm\sqrt\Delta}{2a}\]

  5. Callisto
    • 2 years ago
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    Little correction for the above post... Δ=b^2 - 4ac = (6)^2 - 4(2)(K) < 0 Now you need to solve this: (6)^2 - 4(2)(K) < 0

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