0
To determine whether a polynomial equation has imaginary solutions, you must first identify what type of equation it is. If it is a quadratic equation, you can use the quadratic formula to solve for the solutions. If the equation is a cubic or higher order polynomial, you can use the Rational Root Theorem to determine if there are any imaginary solutions.
The Rational Root Theorem states that if a polynomial equation has rational solutions, they must be a factor of the constant term divided by a factor of the leading coefficient. If there are no rational solutions, then the equation has imaginary solutions.
To use the Rational Root Theorem, first list out all the possible rational solutions. Then, plug each possible rational solution into the equation and see if it is a solution. If there are any solutions, then the equation has imaginary solutions. If not, then there are no imaginary solutions.
Add your answer:
Earn +20 pts
What are the roots of polynomial?
The "roots" of a polynomial are the solutions of the equation polynomial = 0. That is, any value which you can replace for "x", to make the polynomial equal to zero.
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 determine how many solutions a quadratic equation will have and whether they are real or imaginary?
The quadratic equation is: Ax2+ Bx + C = 0-- The equation always has two solutions. They arex = 1/2A [ - B + sqrt(B2 - 4AC) ]x = 1/2A [ - B - sqrt(B2 - 4AC) ] .-- The solutions are real if ( B2 > or = 4AC ).-- The solutions are equal if ( B2 = 4AC ).-- The solutions are complex conjugates if ( B2 < 4AC ).-- The solutions are pure imaginary if ( B = 0 ) & (4AC>0 i.e. -4AC
The discriminant determines how many solutions a quadratic equation will have and whether they are real or?
imaginary
The determines how many solutions a quadratic equation will have and whether they are real or imaginary?
discriminant
Is it possible to have imaginary solutions when solving a polynomial equation?
yes
You can determine by the discriminant whether the solutions to the equation are real or numbers?
imaginary
What are the roots of polynomial?
The "roots" of a polynomial are the solutions of the equation polynomial = 0. That is, any value which you can replace for "x", to make the polynomial equal to zero.
What kind of solution does an equation has if the discriminant is negative?
If the discriminant is negaitve, there are no "real" solutions. The solutions are "imaginary".
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 determine how many solutions a quadratic equation will have and whether they are real or imaginary?
The quadratic equation is: Ax2+ Bx + C = 0-- The equation always has two solutions. They arex = 1/2A [ - B + sqrt(B2 - 4AC) ]x = 1/2A [ - B - sqrt(B2 - 4AC) ] .-- The solutions are real if ( B2 > or = 4AC ).-- The solutions are equal if ( B2 = 4AC ).-- The solutions are complex conjugates if ( B2 < 4AC ).-- The solutions are pure imaginary if ( B = 0 ) & (4AC>0 i.e. -4AC
Without solving the equation determine the nature of the roots of 2x 2 plus 3x plus 9 0?
With a negative discriminant, the two solutions are imaginary.
The discriminant determines how many a quadratic equation will have and whether they are real or imaginary.?
solutions
The discriminant determines how many solutions a quadratic equation will have and whether they are real or?
imaginary
The determines how many solutions a quadratic equation will have and whether they are real or imaginary?
discriminant
If an equation has a degree of three how many solutions will there be?
An equation with a degree of three typically has three solutions. However, it is possible for one or more of those solutions to be repeated or complex.
A quadratic equation can't have one imaginary solution?
That's true. Complex and pure-imaginary solutions come in 'conjugate' pairs.
