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?

## Comments

Sort by:

TopNewest대입해서 복소수상등 무리수상등을 동시해하시면되는데 그게 힘들죠... 증명해보고잇어요! – 찬홍 민 · 2 years, 4 months ago

Log in to reply

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.) – Harsh Shrivastava · 2 years, 6 months ago

Log in to reply

Yes, your observation is correct. Can you figure out how to prove it? – Calvin Lin Staff · 2 years, 6 months ago

Log in to reply