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

Yes, a pretty straightforward question. Is it possible?

  • $\begingroup$ That's the good question. $\endgroup$ – Soner Gönül Jun 16 '12 at 14:58

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 $$x^3-2x=(x-0)\cdot(x^2-2).$$

  • 4
    $\begingroup$ Another simple proof is that if $P$ is a rational degree-2 polynomial with a rational root $a$, then $P/(x-a)$ is a rational degree-1 polynomial: so there is another root and it is rational. $\endgroup$ – Generic Human Jun 16 '12 at 13:38

The equation $$ (x-a)(x-b)=0 $$ is a quadratic equation whose roots are $a$ and $b$. If $a$ is rational and $b$ is irrational then there's the example you seek.

Expanding this, you get $$ x^2 - (a+b)x + ab=0. $$ The coefficient $a+b$ is not rational. Nor is $ab$ unless $a=0$.

If the coefficients are rational then the usual formula for solving quadratic equations tells you that either both roots are rational or both are irrational.


More generally, let $$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0,$$ where $n \ge 1$ and $a_n\ne 0$. Suppose that at least $n-1$ of the (complex) roots of the equation $P(x)=0$ are rational. Then all the roots of $P(x)=0$ are rational if and only if $\frac{a_{n-1}}{a_n}$ is rational. (Here a multiple root is counted according to its multiplicity.)

This is because the sum of the roots is $-\frac{a_{n-1}}{a_n}$.

In particular, if the coefficients are rational and there are at least $n-1$ rational roots, then all roots are rational.

There are polynomials of degree $n$ with exactly $n-1$ rational roots. Use for example $(x-1)(x-2)\cdots(x-(n-1))(x-\sqrt{2})$. Such a polynomial necessarily has at least one irrational coefficient.


Consider the equation of form x(x-a)=0
Here one solution is 0 and other is a. So if a is irrational one root is rational and other is irrational.


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