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A quadratic equation ax2 + bx + c = 0 has the solutions
x = [-b +/- sqrt(b2 - 4*a*c)]/(2*a)
What else can I help you with?
What is the solution of rational equations reducible to quadratic?
They are the solutions for the reduced quadratic.
When the discriminant is perfect square the answer to a quadratic equation will be?
Rational.
Does there exist a quadratic equation whose coefficients are irrational but both the roots are rational?
None, if the coefficients of the quadratic are in their lowest form.
Does there exist a quadratic equation whose coefficients are all distinct irrationals but both the roots are rationals?
Yes, it is possible for a quadratic equation to have distinct irrational coefficients while having rational roots. For example, consider the quadratic equation (x^2 - \sqrt{2}x - \sqrt{3} = 0). The coefficients (-\sqrt{2}) and (-\sqrt{3}) are distinct irrationals, yet the roots of this equation can be rational. Specifically, if we apply the quadratic formula, we can find rational roots depending on the specific values of the coefficients.
Do all rational equations have a single solution?
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.
What is the solution of rational equations reducible to quadratic?
They are the solutions for the reduced quadratic.
When the discriminant is perfect square the answer to a quadratic equation will be?
Rational.
What is the difference with equations with integers and equations with rational numbers?
It is a trivial difference. If you multiply every term in the equation with rational numbers by the common multiple of all the rational numbers then you will have an equation with integers.
Is there a difference between rational expressions and rational equations?
Yes. An equation has an "=" sign.
Does there exist a quadratic equation whose coefficients are irrational but both the roots are rational?
None, if the coefficients of the quadratic are in their lowest form.
Does there exist a quadratic equation whose coefficients are all distinct irrationals but both the roots are rationals?
Yes, it is possible for a quadratic equation to have distinct irrational coefficients while having rational roots. For example, consider the quadratic equation (x^2 - \sqrt{2}x - \sqrt{3} = 0). The coefficients (-\sqrt{2}) and (-\sqrt{3}) are distinct irrationals, yet the roots of this equation can be rational. Specifically, if we apply the quadratic formula, we can find rational roots depending on the specific values of the coefficients.
Do all rational equations have a single solution?
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.
What is true of the discriminant when the two real number solutions to a quadratic equation are rational numbers?
The discriminant must be a perfect square or a square of a rational number.
How can you determine whether a polynomial equation has imaginary 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.
Why do you check your answers for rational equations and radical equations?
You plug the number back into the original equation. If you have a specific example, that would help.
What statement must be true of an equation before you can use the quadratic formula to find the solutions?
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.
Which 5 equations are solving with 4 steps?
To solve equations effectively in four steps, consider these types: Linear Equations: Isolate the variable by adding or subtracting terms, then divide or multiply to solve. Quadratic Equations: Rearrange to standard form, factor or use the quadratic formula, simplify, and solve for the variable. Rational Equations: Clear the denominators, simplify the resulting equation, isolate the variable, and solve. Exponential Equations: Take the logarithm of both sides, isolate the variable, and simplify to find the solution. Systems of Equations: Use substitution or elimination to reduce the system, isolate one variable, and solve for it.
