None, if the coefficients of the quadratic are in their lowest form.
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.
yes * * * * * No it does not. A transcendental number is not rational. It is irrational but, further than that, it is not the root of any polynomial equation with rational coefficients.
Rational.
A quadratic equation ax2 + bx + c = 0 has the solutions x = [-b +/- sqrt(b2 - 4*a*c)]/(2*a)
This is a quadratic expression in x. It cannot be solved because it is an expression: there is no equation to solve. Furthermore, the discriminant for the quadratic = (-10)2 - 4*3*(-5) = 100+60 = 160 which is not a square number. There are, therefore no rational factors. While it would be possible to give irrational factors, factorising a quadratic into linear terms containing irrational numbers is rarely useful.
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.
yes * * * * * No it does not. A transcendental number is not rational. It is irrational but, further than that, it is not the root of any polynomial equation with rational coefficients.
Rational.
The given polynomial does not have factors with rational coefficients.
Real but not a root of an algebraic equation with rational roots coefficients
4.6 is rational.
Rational
An algebraic number is a complex number which is the root of a polynomial equation with rational coefficients.
A quadratic equation ax2 + bx + c = 0 has the solutions x = [-b +/- sqrt(b2 - 4*a*c)]/(2*a)
This is a quadratic expression in x. It cannot be solved because it is an expression: there is no equation to solve. Furthermore, the discriminant for the quadratic = (-10)2 - 4*3*(-5) = 100+60 = 160 which is not a square number. There are, therefore no rational factors. While it would be possible to give irrational factors, factorising a quadratic into linear terms containing irrational numbers is rarely useful.
An algebraic number is one which is a root of a polynomial equation with rational coefficients. All rational numbers are algebraic numbers. Irrational numbers such as square roots, cube roots, surds etc are algebraic but there are others that are not. A transcendental number is such a number: an irrational number that is not an algebraic number. pi and e (the base of the exponential function) are both transcendental.
is 34.54 and irrational or rational. number