Whenever you solve a rational equation you must make sure the result obtained for an answer does not allow the denominator of one of the rational expressions to assume a value of ZERO, as division by zero is undefined and therefore prohibited. For example if we have 2x/(x-3) =(x2 -9x)/ x when we multiply out by x(x-3) we get 2x(x) = (x2 -9x)(x-3) so 2x2 = x(x-9)(x-3) 2x2 = x(x2 - 12x + 27) 2x2 = x3 - 12x2 + 27x so 0 = x3 - 14x2 + 27x 0 = x(x2 - 14x + 27) so solutions are 0 and 7 + √22 and 7 -√22 but 0 makes right hand side expression have zero in denominator so it is not a solution. We actually have to look at all obtained solutions to be sure they ae not extraneous. Suppose we had obtained a 3 for a solution. That would make the left side denominator equal zero and we would have to dismiss that, if 3 was obtained. The two irrational solutions we have obtained are genuine solutions as neither introduces a zero to a denominator
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 is simply an equation with non-rational solutions. There is no special name for it.
The discriminant must be a perfect square or a square of a rational number.
Not all rational equations have a single solution but can have more than one because of having polynomials. All rational equations do have solutions that cannot fulfill the answer.
A rational equation is when its solution can be expressed as a fraction
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 has two complex solutions.
It is simply an equation with non-rational solutions. There is no special name for it.
The discriminant must be a perfect square or a square of a rational number.
Not all rational equations have a single solution but can have more than one because of having polynomials. All rational equations do have solutions that cannot fulfill the answer.
A rational equation is when its solution can be expressed as a fraction
It is important to check your answers to make sure that it doesn't give a zero denominator in the original equation. When we multiply both sides of an equation by the LCM the result might have solutions that are not solutions of the original equation. We have to check possible solutions in the original equation to make sure that the denominator does not equal zero. There is also the possibility that calculation errors were made in solving.
A quadratic equation ax2 + bx + c = 0 has the solutions x = [-b +/- sqrt(b2 - 4*a*c)]/(2*a)
The quadratic formula can be used to find the solutions of a quadratic equation - not a linear or cubic, or non-polynomial equation. The quadratic formula will always provide the solutions to a quadratic equation - whether the solutions are rational, real or complex numbers.
Reciprocal. Except that dividing by a rational equation is much easier.
It is an equation with no solutions [in the given domain]. There may (or may not) be solutions if you change the domain.For example, if X is an integer, then 5X = 2 has no solution. But if you change the domain to rational numbers, then X = 2/5 or 0.4 is a solution.
It is an equation with no solutions [in the given domain]. There may (or may not) be solutions if you change the domain.For example, if X is an integer, then 5X = 2 has no solution. But if you change the domain to rational numbers, then X = 2/5 or 0.4 is a solution.