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The equation is:
x3 - 4x2i - 9xi2+ 36i3 = 0
Note: The ' i ' must be an ordinary constant, and can't be sqrt(-1).
If it were sqrt(-1), then -4i would also be a root of the equation.
Imaginary or complex roots always occur in conjugate pairs.
If -4i is also a root of the equation and the questioner just forgot to include it,
then ' i ' can be sqrt(-1), and the equation can be
x4 + 25x2 + 144 = 0
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Find 2 square roots of 4i?
2i
Can you factor complex numbers?
Yes. Consider, if you can factor complex numbers, then logically, you should be able to take two complex numbers, multiply them together, and get a third. That can indeed be done. For example: (4i + 7)(3i + 2) = -12 + 8i + 21i + 14 = 29i + 2 Therefore, the complex number 29i + 2 must be divisible by 4i + 7 and 3i + 2.
What is 4i plus 55 equal to 2i?
It is a linear equation in a single variable, i. i = -27.5
2 plus 4i - 7 plus 4i?
(2 + 4i) - (7 + 4i) = -5 2 + 4i - 7 + 4i = -5 + 8i
What is a quadratic equation with real coefficents?
A quadratic equation in standard form is based on the model: ax2 + bx + c = 0 a, b, and c are called the coefficients. "Real coefficients" simply means that a, b, and c are real numbers (as opposed to complex numbers). For example: 5x2 - 3x +2 = 0 ... has real coefficients, while (3-2i)x2 + (-2 + 4i)x + (4 + 3i) = 0 does not.
