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Q: How many complex roots does 6x4 plus x3 - 5 equals 0 have?

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No.The roots are the complex conjugate pair 5 Â± 2.4495iwhere i is the imaginary square root of -1.

3x^2 + 8x + 4 = (x+2) (3x+2) x = -2 x = -2/3 So there are no complex roots, they are real. You can test this by b^2 - 4ac if greater than 0, it is real if equal, there will be 2 identical roots. if less than 0 you get imaginary roots.

It is a quadratic equation with one unknown variable, x which has no real roots.

No answer in integers. Quadratic formula gives roots of x2 + x - 26 = 0 as 5.625 and -4.625

The discriminant formula. b2 - 4ac 32 - 4(1)(8) 9 - 32 = - 23 ===========This shows no real roots to this function.

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There are no real root. The complex roots are: [-5 +/- sqrt(-3)] / 2

two complex

one

No real roots. Imaginary roots as this function does not intersect the X axis.

No.The roots are the complex conjugate pair 5 Â± 2.4495iwhere i is the imaginary square root of -1.

The complex numbers (1/2 + √11/2 i) and (1/2 - √11/2 i)

No real roots

The roots are: x = -5 and x = -9

It is a quadratic equation with no real roots or real solutions. In the complex domain, the solutions are 1 +/- i where i is the imaginary square root of -1.

No real roots.

No real roots.

x=16

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