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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.

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  • $\begingroup$ Welcome to Mathematics Stack Exchange. Consider the glb and lub of $\{x\in\mathbb Q|x^2+6x+3<0\}$ $\endgroup$ – J. W. Tanner Aug 2 '20 at 22:01
  • $\begingroup$ Hi, and thanks for the welcome, but I don't understand your comment, I'm new on this $\endgroup$ – MAX SOTO Aug 2 '20 at 22:06
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    $\begingroup$ @MAXSOTO 'glb' means 'greatest lower bound' (the infimum) and 'lub' means 'lowest upper bound' (the supremum). $\endgroup$ – Joe Aug 2 '20 at 23:24
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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

  • each is a non-empty proper subset of $\mathbb {Q} $
  • if a rational number belongs to the set, all the smaller rationals also belong to the set
  • the set has no maximum member ie given any member of the set we can find another member which is greater than the given member

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.

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