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

I have seen that there is a post about this, but I dont understand it. How can it be shown using quadratic formula?

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up vote 10 down vote accepted

Consider $f(x) = (x-p)(x-r)$, where $p \in \mathbb{Q}$ and $r \in \mathbb{R} \backslash \mathbb{Q}$.

If all the coefficients of the quadratic polynomial $ax^2+bx+c$ are rationals, i.e. $a,b,c \in \mathbb{Q}$, then either both the roots are rational or both the roots are irrational. This is because of the fact that the roots are given by $$x^* = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$ Since $a,b,c \in \mathbb{Q}$, the irrational root can occur only when $\sqrt{b^2-4ac}$ is irrational (Why?), i.e., $b^2-4ac \neq (p/q)^2$, where $p,q \in \mathbb{Z}$ and $q \neq 0$. But if $\sqrt{b^2-4ac}$ is irrational say $r$, then both the roots $$\dfrac{-b \pm r}{2a}$$ are irrational. You may want to look at this question to see when the sum of rationals/irrationals is rational/irrational.

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This means that: A quadratic polynomial with rational coefficients cannot have a rational and irrational root? Am I right? – Reader Apr 21 '13 at 19:41
@Reader Yes. Either both roots are rationals or both roots are irrationals. – user17762 Apr 21 '13 at 19:43
Thank you so much. – Reader Apr 21 '13 at 19:44
Can you explain your (why?) in the solution above? – Reader Apr 21 '13 at 20:01
@Reader If $\sqrt{b^2-4ac}$ were to be rational, then $-b \pm \sqrt{b^2-4ac}$ would be rational and so would be $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$. – user17762 Apr 21 '13 at 20:02

Consider $x(x-\pi)=0{}{}{}{}{}{}{}{}{}{}$.

Edit: Apparently the OP wants to consider polynomials only with rational coefficients but he failed to mention so in the question.

Since you're asking for exactly two roots, the polynomial must be something like $\lambda(x-\alpha)(x-\beta)$. Now you want, without loss of generality, $\lambda =1 \land \alpha\in \Bbb Q \land \beta \in \Bbb R\setminus \Bbb Q$.

However $(x-\alpha)(x-\beta)=x^2-(\alpha +\beta)x+\alpha \beta$.

So you wish for $\alpha \beta\in \Bbb Q \land \alpha +\beta\in \Bbb Q$.

But $\alpha \beta\in \Bbb Q \land \alpha +\beta\in \Bbb Q\implies \alpha =0 \land \alpha +\beta\in \Bbb Q \implies \beta \in \Bbb Q$.

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but your polynomial has irrational coefficients. – Reader Apr 21 '13 at 19:28
"How can it be shown using the quadratic formula?" is the question, but this is nevertheless useful. +1 – Pedro Tamaroff Apr 21 '13 at 19:29
@Reader Yes, it does. It that an issue? – Git Gud Apr 21 '13 at 19:29
@Reader Edit it into the question, then. – Git Gud Apr 21 '13 at 19:31
This argument is much easier and also shows that if there is one rational root and one irrational root, then the coefficient of $x$ and the constant terms both must be irrational (assuming the coefficient of $x^2$ is rational). +1 – user17762 Apr 21 '13 at 19:40

Suppose $ax^2+bx+c=a(x-p)(x-q)$ with $a,b,c,p$ rational. Then by expanding and comparing coefficients, you see that $q=-p-\dfrac{b}{a}$. So what can you say about $q$?

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