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

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

How can it be shown using quadratic formula?

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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.

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

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 $q$?