No because the discriminant is less than zero.
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.
The given expression is a quadratic equation. To find its solutions, we can either factor the equation or use the quadratic formula. However, without an equation to solve or any context, it is not possible to provide a numeric answer.
It needs to have an equality or equality signs to have solutions for it.Without any equality signs the given expression can't be considered to be an equation although it might be possible to simplify it.
2
A linear equation is that of a straight line. Any one of the infinitely many points on the line will be solutions. If the equation is in terms of the variables x and y, just pick any two values of x, solve for y and the results will be the coordinates of two solutions.
no
The complex roots of an equation is any solution to that equation which cannot be expressed in terms of real numbers. For example, the equation 0 = x² + 5 does not have any solution in real numbers. But in complex numbers, it has solutions.
Strictly speaking the above equation is a tautological equation or an IDENTITY. An identity is true for all values of any variables that appear in it. Thus, the above "equation" is true for all value of x. - that is, it has infinitely many solutions.
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.
It does not have any solutions! 14.8 is a number, not an equation, inequality or question and so has no solutions.
The given expression is a quadratic equation. To find its solutions, we can either factor the equation or use the quadratic formula. However, without an equation to solve or any context, it is not possible to provide a numeric answer.
It needs to have an equality or equality signs to have solutions for it.Without any equality signs the given expression can't be considered to be an equation although it might be possible to simplify it.
No. If an equation has many solutions, any one of them will satisfy it.
2
There are 0, or no solutions to these equations. If you rearrange each equation to look like an equation of a line, you will have the two lines:y = x -1 and y = x + 1, which are two parallel lines. They do not intersect at any point, so there is no solution.
The solutions of the equation (if any) remain unchanged.
Add the two equations together. This will give you a single equation in one variable. Solve this - it should give you two solutions. Then replace the corresponding variable for each of the solutions in any of the original equations.