Why does a conjugate of a root remain a root of a polynomial?

Question

About quadratic equations, I have the follow question,

What is the polynomial, with rational coefficients, knowing that one root is $ x_1 = 1 + \sqrt 3$ ?

So, to solve the problem, I must know that the other root is $x_2 = 1-\sqrt3$. I see the resolution, then I use $x_2 = 1-\sqrt3$. But I don't know why.

I solved in this way,

$\ \ \ \ \, $ Root Sum $= S = (1+\sqrt3) + (1-\sqrt 3)\, =\, 2$

Root Product $ = P = (1+\sqrt3) * (1-\sqrt3) = -2$

Vieta Formula is $\,x^2 - S\, x + P = 0$ so the polynomial is $\,x^2 - 2\,x - 2 = 0$.

Why is $\,x_2 = 1-\sqrt 3\, $ if $\,x_1 = 1+\sqrt3$?

Why is $\,x_2 = 5+\sqrt 3\, $ if $\,x_1 = 5-\sqrt3$?

Why does this happen (switching the operator of the root) when the coefficients are rational?

Answer 1

If $1+\sqrt{3}$ and $r$ are the roots of your equation, then their sum $r + 1 + \sqrt{3}$ must be rational. So $r=a-\sqrt{3}$ for some rational $a$. Similarly, you can use the fact that their product is rational to conclude that $a$ must be $1$.

Answer 2

Consider Viète's formulas for the coefficients of a polynomial in terms of it's roots. In this specific case, if you have $p(x) = (x - \alpha) (x - \beta) = x^2 - a x + b$, then: $$ a = \alpha + \beta \\ b = \alpha \beta $$ If both $a$ and $b$ are rational, with $\alpha = 1 + \sqrt{3}$, it must be that $\beta = c - \sqrt{3}$ for some $c$ rational. Then: $$ b = \alpha \beta = (1 + \sqrt{3})(c - \sqrt{3}) = c + (c - 1) \sqrt{3} + 3 $$ For this to be rational, $c = 1$.

Answer 3

Hint $ $ Any polynomial with $\rm\color{#0a0}{rational}$ coefficients having root $\,w = 1+\sqrt{3}\,$ also has as root its conjugate $\,\bar w = 1-\sqrt{3}\,$ since, in the field $\,\rm F =\color{#0a0}{\Bbb Q}(\sqrt 3) = \{ a\!+\!b\sqrt 3\, :\, a,b\in\Bbb Q\}\,$ we have

Key Idea $ $ Conjugation $\rm\:w=a+b\sqrt{3}\,\mapsto\, \bar w = a-b\sqrt{3}\in F\:$ $\rm\:\color{#c00}{preserves\ sums\,\ \&\,\ products}.\:$ Further, conjugation $\rm\color{#0a0}{fixes\ rationals}\in\color{#0a0}{\Bbb Q}.\:$ Hence, by induction, it preserves polynomial functions with $\rm\color{#0a0}{rational}$ coefs, i.e. $\rm\ \overline{f(w)} = f(\overline w),\,$ for all $\rm\, f(x)\in\color{#0a0}{\Bbb Q}[x],\, $ since such polynomials are compositions of said basic sum & product operations. $ $ More explicitly $$ \begin{eqnarray} \rm \overline{f(w)}\: &=&\rm\ \ \overline{a_n w^n +\,\cdots + a_1 w + a_0}\\ &=&\rm\,\ \overline{a_n w^n}\, +\,\cdots + \overline{a_1 w} + \overline a_0\quad by\ \ \ \color{#c00}{\overline{x+y}\, =\, \overline x + \overline y}\ \ \ \forall\ x,y \in F\\ &=&\rm\,\ \overline a_n\, \overline w^n+\,\cdots + \overline a_1\overline w + \overline a_0\quad by\ \ \ \color{#c00}{\overline{x\, *\, y}\, =\, \overline x\:\! *\, \overline y}\ \ \ \forall\ x,y \in F \\ &=&\rm\,\ a_n\, \overline w^n + \,\cdots + a_1 \overline w + a_0\quad by\ \ \ \color{#0a0}{\overline a = a}\ \ \forall\ \color{#0a0}a\in \color{#0a0}{\Bbb Q}\\ &=&\rm\ f(\overline w)\\ \rm\!\! So\ \ 0\! =\! f(w) \Rightarrow \bar 0 = \overline{f(w)}& =&\ \rm f(\overline w),\ \ i.e.\ \ \bbox[6px,border:2px solid #c00]{w\ root\ of\ f\,\Rightarrow\, \overline w\ root\ of\ f}\quad {\bf QED} \end{eqnarray}$$

This usually fails if $\rm\,f\,$ has coefficients $\color{#0a0}{\not\in\Bbb Q}\,$ e.g. $\rm\,\bar w\,$ is a root of $\rm\,x\!-\!w\,$ iff $\rm\,\bar w = w,\,$ i.e. $\rm\,w\in \Bbb Q.$

Remark $ $ The analogous polynomial preservation property holds true for any algebraic structure, i.e. since homomorphisms preserve the basic operations (including constants = $0$-ary operations), they also preserve the "polynomial" terms composed of these basic operations. Said equivalently, hom's commute with polynomials.

Answer 4

You can see this answer as a supplement to Bill Dubuque's, but I am avoiding any technical language.

The key to understanding why if $1+\sqrt{5}$ is a root of a quadratic equation with rational coefficients, then $1-\sqrt{5}$ is also a root, is understanding the symmetry between $\sqrt{5}$ and $-\sqrt{5}$. Here is what I mean:

If you write down any expressions with rational numbers and $\sqrt{5}$, and you do math with them, then all the math is dictated by the fact that $\sqrt{5}$'s square is 5. Example:

$$(1+\sqrt{5})(3-2\sqrt{5}) = 3 - 2\sqrt{5} + 3\sqrt{5} - 2\cdot (\sqrt{5})^2=3+\sqrt{5} - 10 = -7 + \sqrt{5}$$

You get the answer by doing purely rational arithmetic except occasionally whenever you see $(\sqrt{5})^2$, you can write $5$. In other words, the mathematical behavior of $\sqrt{5}$ in this context is completely determined by the fact that $\sqrt{5}$ satisfies the equation $x^2=5$.

Now the key point is this: $-\sqrt{5}$'s square is also 5. In other words, it also satisfies the equation $x^2=5$. This means that $-\sqrt{5}$ behaves in the exact same way as $\sqrt{5}$ does. If we replace $\sqrt{5}$ with $-\sqrt{5}$ throughout the calculation above, everything will work the exact same way, and $-\sqrt{5}$ will also replace $\sqrt{5}$ in the answer. Look:

$$(1-\sqrt{5})(3-2(-\sqrt{5})) = 3 -2(-\sqrt{5}) + 3(-\sqrt{5}) - 2\cdot(-\sqrt{5})^2 $$

$$= 3 - \sqrt{5} - 10 = -7-\sqrt{5}$$

Replacing $\sqrt{5}$ with $-\sqrt{5}$ in the problem just replaced it as well throughout the calculation and in particular in the answer.

So, to your original question, if $1+\sqrt{5}$ is a root of a quadratic equation $ax^2+bx+c=0$ with rational coefficients, that means that

$$a(1+\sqrt{5})^2 + b(1+\sqrt{5})+c=0$$

But by the above train of thought, if you replace $\sqrt{5}$ with $-\sqrt{5}$ throughout the calculation, everything should still work the same. Since you are assuming that $a,b,c$ are rational, then they don't have any $\sqrt{5}$ in them, so there is nothing to replace, and it looks like

$$a(1-\sqrt{5})^2 + b(1-\sqrt{5}) + c = 0$$

So $1-\sqrt{5}$ is also a root.