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?
$\begingroup$
$\endgroup$
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
5
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.
-
$\begingroup$ This means that: A quadratic polynomial with rational coefficients cannot have a rational and irrational root? Am I right? $\endgroup$– ReaderCommented Apr 21, 2013 at 19:41
-
$\begingroup$ @Reader Yes. Either both roots are rationals or both roots are irrationals. $\endgroup$– user17762Commented Apr 21, 2013 at 19:43
-
$\begingroup$ Can you explain your (why?) in the solution above? $\endgroup$– ReaderCommented Apr 21, 2013 at 20:01
-
1$\begingroup$ @Reader If $\sqrt{b^2-4ac}$ were to be rational, then $-b \pm \sqrt{b^2-4ac}$ would be rational and so would be $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$. $\endgroup$– user17762Commented Apr 21, 2013 at 20:02
-
$\begingroup$ Maybe a better way to state the reason: The sum of the roots is $-b/a$, which is rational. But the sum of a rational plus an irrational must be irrational. A more elementary reason than the quadratic formula. $\endgroup$– GEdgarCommented Aug 21, 2013 at 12:44
$\begingroup$
$\endgroup$
6
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 $\alpha \beta\in \Bbb Q \land \alpha +\beta\in \Bbb Q\implies \alpha =0 \land \alpha +\beta\in \Bbb Q \implies \beta \in \Bbb Q$.
-
$\begingroup$ but your polynomial has irrational coefficients. $\endgroup$– ReaderCommented Apr 21, 2013 at 19:28
-
$\begingroup$ "How can it be shown using the quadratic formula?" is the question, but this is nevertheless useful. +1 $\endgroup$– Pedro ♦Commented Apr 21, 2013 at 19:29
-
1$\begingroup$ @Reader Yes, it does. It that an issue? $\endgroup$– Git GudCommented Apr 21, 2013 at 19:29
-
2$\begingroup$ @Reader Edit it into the question, then. $\endgroup$– Git GudCommented Apr 21, 2013 at 19:31
-
1$\begingroup$ 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 $\endgroup$– user17762Commented Apr 21, 2013 at 19:40
$\begingroup$
$\endgroup$
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 $q$?
answered Apr 21, 2013 at 19:34