As there is no system of equations shown, there are zero solutions.
3
There are infinitely many ordered pairs tat are solutions. They are all points on the line represented by 5x-6y = 13
It has the following solutions.
A unit circle is in the coordinate plane where both axes are measured in real numbers. The imaginary circle is in the complex plane in which one axis (horizontal) measures the real component of a complex number and the other axis measures the imaginary component.
That's true. Complex and pure-imaginary solutions come in 'conjugate' pairs.
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.
As stated in the attached link, there are three possible discriminant conditions: Positive, Zero, or Negative. If the discriminant is negative, there are no real solutions but there are two imaginary solutions. So, yes there are solutions if the discriminant is negative. The solutions are imaginary, which is perfectly acceptable as solutions.
Which 2 order pairs are solutions of y x + 5
If the discriminant is negaitve, there are no "real" solutions. The solutions are "imaginary".
No. A quadratic may have two identical real solutions, two different real solutions, ortwo conjugate complex solutions (including pure imaginary).It can't have one real and one complex or imaginary solution.
yes
The cast of The Imaginary Solutions of Thomas Chimes - 2007 includes: Thomas Chimes as himself Phillip Mitsis as himself
you can have either one or three x-intercepts, but now 2. because two real roots means 1 imaginary root which is not possible since imaginary roots come in pairs (2,4,6,8...)
That depends on the equation.
imaginary
You cannot make your imaginary friend come to life.