# Dedekind Cuts to solving quadratic equations

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.

• Welcome to Mathematics Stack Exchange. Consider the glb and lub of $\{x\in\mathbb Q|x^2+6x+3<0\}$ Aug 2, 2020 at 22:01
• Hi, and thanks for the welcome, but I don't understand your comment, I'm new on this Aug 2, 2020 at 22:06
• @MAXSOTO 'glb' means 'greatest lower bound' (the infimum) and 'lub' means 'lowest upper bound' (the supremum).
– Joe
Aug 2, 2020 at 23:24

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 $$B=\{x\mid x\in\mathbb {Q}, x<-3,x^2+6x+3>0\}$$
• each is a non-empty proper subset of $$\mathbb {Q}$$
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.