I'm trying to prove the following, which is left unproven in something I'm reading on ruler-and-compass constructions:(adsbygoogle = window.adsbygoogle || []).push({});

If [itex]ax^3+bx^2+cx+d[/itex] is a polynomial over a subfield F of ℝ, and [itex]p+q\sqrt{r}[/itex] is a root (with [itex]\sqrt{r}\notin F[/itex]) then [itex]p-q\sqrt{r}[/itex] is also a root.

The theorem immediately before this made use of expanding [itex]a(x-r_1)(x-r_2)(x-r_3)[/itex] and equating coefficients to get a system of three equations in the roots and coefficients. The book suggested that this proof was similar, but I was't able to derive anything useful from that.

I also tried defining [itex]g(t)=f(p-t\sqrt{r})-f(p+t\sqrt{r})[/itex], and noting that [itex]g(q)=f(p-q\sqrt{r})[/itex] and [itex]-g(-t)=g(t)[/itex].

Then we see that

[itex]f(p-q\sqrt{r})=g(q)=-g(-q)=-f(p+q\sqrt{r})=0[/itex].

However, I know this must be flawed, because I don't believe I ever used the assumption that [itex]\sqrt{r}\notin F[/itex]. Any thoughts?

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# Roots of Cubic Polynomials over R

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