No.The roots are the complex conjugate pair 5 ± 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.
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 ± 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.
5n2 + 2n + 6 = 0 is a quadratic equation in the variable n.The equation does not have any real roots. The roots are the complex conjugate pair -0.2 ± 1.077i where i is the imaginary square root of -1.
No real roots.
3x^2=-20 so x^2=-20/3 this has no real roots but the plus of minus i(square root of 20/3) is a solution if we allow complex numbers.