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.

It sounds like you know how to solve $ax^2+bx+c=0$. Without doing any extra work, your problem has $a=1$, $b=-2m$, $c=-2m-1$. Can you finish it from here? — Alex R., Aug 21 '15 at 20:46
$x^2-2mx-2m-1=x^2+((-2m-1)+(1))\times x + (1)\times (-2m-1)=(x-2m-1)(x+1)$ — Math-fun, Aug 21 '15 at 20:59
You should have $x^2-2mx\color{red}{+}(2m+1)(-1)=0,$ so you are meant to observe that the roots are $2m+1$ and $-1$ rather than use the formula — David Quinn, Aug 21 '15 at 21:26
@DavidQuinn, to use this method is mandatory that the cofficient of $x^2$ is $1$, right? — Valeria Gonzales, Aug 21 '15 at 22:24

score 3 · Accepted Answer · answered Aug 21 '15 at 20:46

3

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

answered Aug 21 '15 at 20:46

juantheron

53,015

score 0 · Answer 2 · answered Aug 21 '15 at 20:49

0

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

answered Aug 21 '15 at 20:49

Dr. Sonnhard Graubner

95,283

score 0 · Answer 3 · answered Aug 21 '15 at 20:50

0

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 $(2m+1)(-1)$

answered Aug 21 '15 at 20:50

David Quinn

34,121

Thanks, it's interesting to see how you make: $$(-2m - 1)$$ to $$(2m + 1)(-1)$$ May you explain it? – Valeria Gonzales Aug 21 '15 at 20:57
By removing the common factor of $-1$. – David Quinn Aug 21 '15 at 20:59
And why its necessary to do so? I'm not seeing the big picture. – Valeria Gonzales Aug 21 '15 at 21:00
It just happens that with this example and the previous example you posted, we can reduce the quadratic in this way without having to resort to the formula or completing the square. Are you perhaps working on a set of questions where this is the theme of the exercise? – David Quinn Aug 21 '15 at 21:03
Yes, sort of. Please David, see my update in the question. This page is such a great help, since we can't afford a private tutor. – Valeria Gonzales Aug 21 '15 at 21:17

score 0 · Answer 4 · answered Aug 21 '15 at 22:05

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.

Jon · Answer 5 · 2015-08-21T22:13:02.863

0

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...

edited Aug 21 '15 at 22:13

answered Aug 21 '15 at 22:06

Jon

1

If $x=-1$ your solution divides by zero. Hope that helps. – Karl Aug 21 '15 at 22:25

Better way of solving this quadratic equation?

5 Answers5