Better way of solving this quadratic equation?

Question

This is the excersice:

$$x^2 - 2mx - 2m - 1 = 0$$

I've done this:

1.- $$x^2 - (2mx -2m) - 1 = 0$$ 2.- $$x^2 - (2m(x-1)) - 1 = 0$$

I think I need to transform:

$$(2m(x-1))$$ to something of the form $$mx$$

And then to have something like:

$$x^2 - mx - 1 = 0$$

So after I can apply the quadratic formula, but not sure how to proceed.

UPDATE 1:

After David answer, i have this:

$$x^2 - 2mx - (2m + 1)(-1) = 0$$

So: a -> 1, b -> -2m, and c -> (2m + 1)(-1).

If it is correct, is right to assume that the answer marked as correct from "juantheron" is more practical?

I see the solution on the marked answer, but I want to understand every other possibilities.

Answer 1

Given $x^2-2mx-2m-1 = 0\Rightarrow x^2-2mx+m^2 = m^2+2m+1$

So we get $(x-m)^2 = (m+1)^2\Rightarrow (x-m) = \pm(m+1)$

So we get $x = m\pm (m+1) = 2m+1,-1$

Answer 2

we get $x_{1,2}=m\pm\sqrt{m^2+2m+1}$ and we get $$x_1=m+m+1=2m+1$$ or $$x_2=m-m-1=-1$$

Answer 3

See my comments to your previous question. You can use the formula, or you can observe that the sum of roots is $2m$ and the product of roots is $(2m+1)(-1)$

Answer 4

If there are to be solutions in $\mathbb{R}$ then the discriminant $\Delta$ should be none negative or $\Delta\ge0$. This requires $(-2m)^2-4(-2m-1) \ge 0\implies 4m^2+8m+4\ge0$ This can be written as $(m+1)^2\ge0$ and is true for all $m$.

As the coefficient of $x$ is even completing the square would work out nicely. $$(x-m)^2-m^2-2m-1=0\\(x-m)^2=(m+1)^2\\x=\pm \sqrt{(m+1)^2}+m$$

Hope this helps.

Answer 5

Well it's already answered, but since you asked for other ways too

$$x^2-2mx-2m-1=0$$ Moving things around: $$x^2-1=2mx+2m$$ By difference of squares on LHS and factoring out $2m$ on RHS: $$(x-1)(x+1)=2m(x+1)$$ $$(x-1)(x+1)/(x+1)=2m$$

$$x-1=2m$$ therefore we end up with, $$x=2m+1$$ I seem to be missing the $x=-1$ though...