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If 3 - 2i is a solution to a polynomial equation is the complex conjugate 3 plus 2i also a solution yes or no?
Wiki User
∙ 15y ago
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.
Wiki User
∙ 15y ago
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Can you find a third degree polynomial equation with rational coefficients that has the given numbers as roots 3i and 7?
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
If a and b are any real numbers what is the conjugate of a plus b?
The concept of conjugate is usually used in complex numbers. If your complex number is a + bi, then its conjugate is a - bi.
What is the conjugate of 11 plus 5i?
To find the complex conjugate change the sign of the imaginary part: For 11 + 5i the complex conjugate is 11 - 5i.
A polynomial has one root the equals 4 plus 17i name one other root of this polynomial?
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.
What is the complex conjugate of 2 plus 3i?
[ 2 - 3i ] is.
A quadratic equation can't have one imaginary solution?
That's true. Complex and pure-imaginary solutions come in 'conjugate' pairs.
What is the value of x in the equation x squared minus x plus 7 equals 3?
The equation does not have a real number solution. Using the quadratic formula will give it's conjugate pair complex solution.
Can a polynomial equation have no answer?
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.
What solution does a negative discriminant have?
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.
The graph of a polynomial changes direction twice and has only one root What can you say about the polynomial?
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.
What is an algebraic number?
An algebraic number is a complex number which is the root of a polynomial equation with rational coefficients.
Which polynomial has rational coefficients a leading leading coefficient of 1 and the zeros at 2-3i and 4?
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.
How do you use the discriminant to find the number and type of zeros of a function?
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.
Can you find a third degree polynomial equation with rational coefficients that has the given numbers as roots 3i and 7?
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
What is the solution set of the quadratic equation 8x2 plus 2x plus 1 0?
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.
Complex conjugate of 5i-4?
To get the complex conjugate, change the sign in front of the imaginary part. Thus, the complex conjugate of -4 + 5i is -4 - 5i.
What is the complex conjugate of 2-3i?
The complex conjugate of 2-3i is 2+3i.
