Let t be a transcendental number. Prove that t cannot be a root of any equation of the form x^2+ax+b=0, where a and b are constructible numbers.

Question

Let t be a transcendental number. Prove that t cannot be a root of any equation of the form x^2+ax+b=0, where a and b are constructible numbers. Hint: you can use the fact that the constructible numbers are algebraic

i feel confused about this question, hope someone could give me some ideas.

Answer 1

Suppose not. Let $K$ be the field obtained by adjoining $a$ and $b$ to the rationals. Since $K$ is obtained by adjoining a finite number of algebraic numbers, $K$ is a finite extension of the rationals. Let $L$ be the field obtained by adjoining $t$ to $K$. The $L$ is of degree 2 over $K$; indeed, a basis for $L$ as a vector space over $K$ is given by $\{\,1,t\,\}$, since $t^2=-at-b$ is a linear combination of $1$ and $t$, as are all higher powers of $t$. By the Tower Law, $L$ is finite over the rationals, so algebraic over the rationals, so every element of $L$ is algebraic, contradiction to $t$ being transcendental.

Answer 2

If a number $x$ satisfies $x^2+ax+b$, with $a,b$ algebraic then just solve for $x$! $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} $$

Now algebraic numbers are a field (Specifically closed under $+,\times,/,-$). Also if $a$ is algebraic $\sqrt{a}$ is too (Proof: If $a$ satisfies a polynomial $f(x)=0$, the $\sqrt{a}$ satisfies $f(x^2)=0$), it follows that $x$ is algebraic too!