[Math] Is it possible for a quadratic equation to have one rational root and one irrational root

Question

Is it possible for a quadratic equation to have one rational root and one irrational root?

Yes, a pretty straightforward question. Is it possible?

Answer 1

Yes; for example, the quadratic equation $$x^2-\sqrt{2}x=0$$ has two solutions, namely $x=0$ (which is rational) and $x=\sqrt{2}$ (which is irrational).

However, if we only allow rational coefficients for our quadratic equation, then it is true that either both solutions are rational or both are irrational. Given rational numbers $a$, $b$, and $c$, with $a\neq0$, the quadratic equation tells us that the solutions to $ax^2+bx+c=0$ are $$x=\frac{-b+\sqrt{b^2-4ac}}{2a},\quad x=\frac{-b-\sqrt{b^2-4ac}}{2a}.$$ If $\sqrt{b^2-4ac}$ is irrational, then both of these numbers are irrational; if $\sqrt{b^2-4ac}$ is rational, then both of these numbers are rational.

This property is particular to quadratic equations only. For example, the cubic equation $$x^3-2x=0$$ has the solutions $x=0$, $x=\sqrt{2}$, and $x=-\sqrt{2}$, the first of which is rational and the latter two of which are irrational, even though all of its coefficients are themselves rational numbers. The explanation for this is that a quadratic polynomial over the rational numbers must either factor completely or be irreducible, while a higher-degree polynomial can factor partially. For example, the factorization of $x^3-2x$ into irreducible rational polynomials is $$x^3-2x=(x-0)\cdot(x^2-2).$$