Not necessarily, take for example the equation x^2=5-12i. Then, 3-2i satisfies the equation. However, 3+2i does not because (3+2i)^2 = 5+12i.
Yes, easily. Even though the question did not ask what the polynomial was, only if I could find it, here is how you would find the polynomial: Since the coefficients are rational, the complex (or imaginary) roots must form a conjugate pair. That is to say, the two complex roots are + 3i and -3i. The third root is 7. So the polynomial, in factorised form, is (x - 3i)(x + 3i)(x - 7) = (x2 + 9)(x - 7) = x3 - 7x2 + 9x - 63
The concept of conjugate is usually used in complex numbers. If your complex number is a + bi, then its conjugate is a - bi.
To find the complex conjugate change the sign of the imaginary part: For 11 + 5i the complex conjugate is 11 - 5i.
There is not enough information. You can't calculate one root on the basis of another root. HOWEVER, if we assume that all the polynomial's coefficients are real, then if the polynomial has a complex root, then the complex conjugate of that root (in this case, 4 - 17i) must also be a root.
Well, honey, a polynomial equation can have multiple solutions, depending on the degree of the polynomial. A polynomial of degree "n" can have at most "n" solutions, including complex solutions. So buckle up and solve those equations, darlin'!
That's true. Complex and pure-imaginary solutions come in 'conjugate' pairs.
The equation does not have a real number solution. Using the quadratic formula will give it's conjugate pair complex solution.
To efficiently solve complex polynomial equations using the Wolfram Polynomial Calculator, input the polynomial equation you want to solve into the calculator. Make sure to include all coefficients and variables. The calculator will then provide you with the solution, including real and complex roots, if applicable. You can also adjust the settings to customize the output format and precision of the results.
It depends on the discriminant value of the quadratic equation. If the discriminant is positive, there are two distinct real solutions; if it is zero, there is one real solution; and if it is negative, there are two complex conjugate solutions.
It depends on the domain. In the complex domain, a polynomial of order n must have n solutions, although some of these may be multiple solutions. In the real domain, a polynomial of odd order must have at least one real solution, while a polynomial of even order may have no real solutions.
In basic mathematics, a quadratic equation with a negative discriminant has no solutions. However, at a more advanced level you will learn that it has two solutions which form a complex conjugate pair.
It is a polynomial of odd power - probably a cubic. It has only one real root and its other two roots are complex conjugates. It could be a polynomial of order 5, with two points of inflexion, or two pairs of complex conjugate roots. Or of order 7, etc.
There cannot be such a polynomial. If a polynomial has rational coefficients, then any complex roots must come in conjugate pairs. In this case the conjugate for 2-3i is not a root. Consequently, either (a) the function is not a polynomial, or (b) it does not have rational coefficients, or (c) 2 - 3i is not a root (nor any other complex number), or (d) there are other roots that have not been mentioned. In the last case, the polynomial could have any number of additional (unlisted) roots and is therefore indeterminate.
An algebraic number is a complex number which is the root of a polynomial equation with rational coefficients.
If the discriminant is positive, then the function has two real zeros. If it is zero, then the function has one real zero. If it is negative, then it has two complex conjugate zeros.This assumes that we are talking about a standard second order polynomial equation, i.e. quadratic equation, in the form Ax2 + Bx + C = 0, and that the discriminant is B2 - 4AC, which is a part of the standard solution of these kind of equations.
Yes, easily. Even though the question did not ask what the polynomial was, only if I could find it, here is how you would find the polynomial: Since the coefficients are rational, the complex (or imaginary) roots must form a conjugate pair. That is to say, the two complex roots are + 3i and -3i. The third root is 7. So the polynomial, in factorised form, is (x - 3i)(x + 3i)(x - 7) = (x2 + 9)(x - 7) = x3 - 7x2 + 9x - 63
There is no equals sign visible but, taking the equation to be8x2 + 2x + 1 = 0,the solutions are the complex conjugate pair, -1.25 ± 0.330719*i where i is the imaginary square root of -1.