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# Just Curious

As we all know, if $$a_0+a_1x+...+a_nx^n=0$$ has a solution of $$p+qi$$ where $$\left\{ a_0,a_1,...,a_n \right\} \subset \mathbb{R}\$$, then $$p-qi$$ is also a solution.

Similarly, if $$a_0+a_1x+...+a_nx^n=0$$ has a solution of $$p+q\sqrt { m }$$ where $$\left\{ a_{ 0 },a_{ 1 },...,a_{ n } \right\} \subset \mathbb{Q}\ \quad and\quad \sqrt { m } \in \mathbb{I}\$$, then $$p-q\sqrt { m }$$ is also a solution.

I was playing around with it until I suddenly got curious. If $$a_{ 0 }+a_{ 1 }x+...+a_{ n }x^{ n }=0\quad (n\ge 4\quad and\quad n\in \mathbb{N}\ )$$ has a solution of $$p\sqrt { m } +qi\quad (\sqrt { m } \in \mathbb{I}\ )$$, then is it safe to assume that it also has the solutions of $$p\sqrt { m } -qi,\quad -p\sqrt { m } +qi,\quad -p\sqrt { m } -qi\quad$$?

I tried it for $$n=4$$, and it worked.

Can someone confirm this for all values of n's?

Note by Nick Lee
3 years, 2 months ago

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- 3 years ago

Let $$p + q\sqrt{m}$$ be a root of a polynomial $$f(x)$$.

We can write $$f(x)$$ = $$(x-(p + q\sqrt{m})$$$$(x-(p - q\sqrt{m}))$$$$a(x)$$ + $$r(x)$$

$$r(x)$$'s degree would be 1,

$$\implies r(x) = ax + b$$

If we get $$a=b=0$$,then $$p - q\sqrt{m}$$ will also be root of $$f(x)$$

Put $$x = p + q\sqrt{m}$$ & get $$a=b=0$$.

(This is not a proof but a hint.)

- 3 years, 2 months ago

Yes, your observation is correct. Can you figure out how to prove it?

Staff - 3 years, 2 months ago

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