I need some help to solving a quadratic equation, but translating it into Dedekind Cuts and using the completing the square way. I first solved the equation in a usual way so i can know where do i have to get, but the Dedekind Cut stuff, I don't know how to use it on it.
The equation is: $$x^2 + 6 x + 3 = 0$$
I know that: $$x_{1} = -\sqrt{6}-3$$ and $$x_{2} = -3+\sqrt{6}$$ but I have to found if there is any solution on $\Bbb R$
I'm not sure if this question qualifies as Real Analysis; I'm sorry.
You can notice that one of the roots is less than $-3$ and other one is greater than $-3$ and the sign of $x^2+6x+3$ is negative between these roots. So for the greater root use $$A=\{x\mid x\in\mathbb {Q}, x>-3,x^2+6x+3<0\}\cup\{x\mid x\in\mathbb{Q}, x\leq - 3\}$$ And for the other root use $$B=\{x\mid x\in\mathbb {Q}, x<-3,x^2+6x+3>0\}$$
Confirm that both these sets satisfy the following defining properties of a Dedekind cut
And then you need to show further that these sets $A, B$ indeed satisfy the equation $x^2+6x+3=0$. This will require you to know how to multiply and add Dedekind cuts. This part of the exercise is boring and lengthy.
The overall exercise can be made a little simpler if one rewrites the equation as $(x+3)^2=6$ (completing the square) and you may use this aspect in the above proofs.