Would we classify a quadratic expression as the following as a polynomial?


Shouldn't the roots of a quadratic expression always occur in pairs of either rational or irrational expressions?

  • 3
    $\begingroup$ No, that's only true if the coefficients are rational. They always are for the first couple of years because ther's no point making things complicated. $\endgroup$
    – Empy2
    Sep 21 '18 at 15:56
  • $\begingroup$ So would we not consider this expression a polynomial? Coefficients of all the terms need to be rational or the leading term? @Empy2 $\endgroup$ Sep 21 '18 at 16:46
  • 4
    $\begingroup$ It's a quadratic polynomial over the reals, but not over the rationals. $\endgroup$
    – Ned
    Sep 21 '18 at 16:48

If the coefficients $a,b,c$ in the equation $ax^2+bx+c=0$ are rational and $a\ne0,$ then the roots are $$\left( \dfrac{-b}{2a} \pm (\text{a number that may be rational or irrational}) \right).$$ In that case either both roots are rational or both are irrational. But if some of $a,b,c$ are not rational, then that conclusion does not hold.

And if $a,b,c$ are real then either both roots are real or both have a nonzero imaginary part, and the two imaginary parts cancel each other out if added. But if the quadratic equation is $x^2 -(5+4i)x +(11+7i)=0,$ then the roots are not complex conjugates of each other. (For that equation, the discriminant $b^2-4ac$ is $-35+12i$ and $\displaystyle \pm\sqrt{b^2-4ac\,\,} = \pm(1+6i).$)

  • $\begingroup$ You might want to clarify to what question the answer is "No": there are three questions (one in the title and two in the body) in the OP. $\endgroup$
    – NickD
    Sep 21 '18 at 19:27
  • $\begingroup$ The 'No' is to the question regarding the roots being a rational pair and an irrational pair. $\endgroup$ Sep 24 '18 at 3:26

The expression you wrote is a polynomial in the variable $s$ over the real numbers $\mathbb{R}$. It is an expression of this type: \begin{equation} a_0+a_1x+a_2x^2+\dots+a_nx^n \end{equation} where $a_i \in \mathbb{R} $ for all $i, 0 \le i \le n$.

Of course you can choose to consider polynomials over the rational numbers, or over the integers; in those cases, the expression you wrote is not a polynomial over the rationals or the integers because not all the coefficients are in $\mathbb{Q}$ or in $\mathbb{Z}$.

Let's now talk about your last question: the root of a quadratic polynomial over the real numbers (we can include the rationals and the integers as special cases of real numbers) are either both imaginary or both real, but nothing is said a priori without knowing the coefficients about them. You can have a quadratic polynomial over the real numbers whose roots are an integer and an irrational number: \begin{equation} x^2-(1+\sqrt{2})x+\sqrt{2}=(x-1)(x-\sqrt{2}). \end{equation}

  • 2
    $\begingroup$ Good answer. +1. $\endgroup$
    – Lubin
    Sep 21 '18 at 21:26

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