There is nothing TECHNICALLY wrong that I see in this revision.
However, I am concerned about whether you understand why one step is both necessary and correct. You go
[MATH]x^2 + \dfrac{bx}{a} = -\ \dfrac{c}{a} \implies \left ( x + \dfrac{b}{2a} \right )^2 - \left ( \dfrac{b}{2a} \right)^2 = -\ \dfrac{c}{a}[/MATH],
which is true only if
[MATH]x^2 + \dfrac{bx}{a} = \left ( x + \dfrac{b}{2a} \right )^2 - \left ( \dfrac{b}{2a} \right )^2.[/MATH],
That crucial step is true, but you do not show why it is true. In other words, you have skipped the most important step in the proof. That worries me because it implies that you have memorized something rather than understood it.
[MATH]\text {Let } u = \dfrac{b}{2a} \text { and } v = -\ \dfrac{c}{a}.[/MATH]
[MATH]\therefore x^2 + \dfrac{bx}{a} = -\ \dfrac{c}{a} \implies[/MATH]
[MATH]x^2 + \left ( \dfrac{b}{a} * \dfrac{2}{2} * x \right ) = v \implies[/MATH]
[MATH]x^2 + \left ( 2 * \dfrac{b}{2a} * x \right ) = v \implies[/MATH]
[MATH]x + 2wx = v \implies[/MATH]
[MATH]x + 2wx + w^2 - w^2 = v \implies[/MATH]
[MATH]x + 2wx + w^2 = w^2 + v \implies[/MATH]
[MATH] (x + w)^2 = w^2 + v \implies[/MATH]
[MATH]x + w = \pm \sqrt{w^2 + v} \implies[/MATH]
[MATH]x = -\ w \pm \sqrt{w^2 + v} \implies[/MATH]
[MATH]x = -\ \dfrac{b}{2a} \pm \sqrt{\left ( \dfrac{b}{2a} \right )^2 - \dfrac{c}{a}}.[/MATH]
That is the crux of the proof. Do you understand it?
Everything else is just simplification. The insight is in what is shown.
Hi Jeff, I see you first made it into a Perfect square trinomial after which you closed it into a square binomial
(x+b/2a)² by which point you already had the (b/2a)² positive on the RHS. After adding the extra (b/2a)² negative from the LHS.
Here is what I understand:
Binomial square (a+b)²
=
Trinomial square a² +2ab + b²
Say you have x²+6x. The equivalent to this will be (x+3)² -9. How?
Because when it is expanded, the middle term will be twice the value i.e. 2ab. You will then have an extra value i.e. b² too in the trinomial, so to make the two equivalent (by two I mean the (x+3)² to the (x² + 6x) ) you have to remove the extra square i.e. b² of the trinomial which would be 3² in this case which is 9. This removal makes the value equivalent
So (x+3)² - 9 = x² + 6x
Which is really:
x² +3x +3x +9 - 9 = x² + 6x
The coefficient of x is 1
We can apply this known fact of (a+b)² as you explained
Since we know that 3 is half of the original 6x, we simply halve the b/a in our equation by 2, we also know that we have to minus the extra square at the end for equivalence like the 9 in the previous equation.
So then based on this, we simply make the equation thus:
(x+b/2a)² - (b/2a)²
The 2a is there in the second term " - (b/2a)² " because we removed the "9" in the trinomial AFTER x²+ 6x became (x+3)² i.e. it had already halved and since b/a is 6x and the 3 in (x+3)² can be said to be b/a * ½, in context, so since we half the 6 and then square the 3 AND THEN remove it, similarly, we half (b/a) , square it AND THEN remove it, so it's (b/2a)² that we remove.
Therefore:
(x+b/2a)² - (b/2a)² = - c/a
Then to further balance we remove the minus which means adding (b/2a)² to both sides so we thus have a positive version of the b² trinomial i.e. a positive version of - (b/2a)² if you will.
All we've done then is make x² + 6x into (x+3)² - 9. It is simply a conversion.
And so in the context of our equation we say:
x² + 6*x
x² + b/a*x
Becomes it's equivalent
(x+3)² - 9
(x+b/2a)² - (b/2a)²
We then remove the negative from LHS which means adding the same to both sides.
So then:
(x+b/2a)² - (b/2a)² = - c/a
Becomes:
(x+b/2a)² = - c/a + (b/2a)²
(x+b/2a)² = - c/a + (b)²/(2a)²
Then we continue forth from this.
P.s.
Above, you haven't squared the x, I assume that was a typing error, right?
x +2wx = v
x +2wv +w² -w² = v
And Can you clarify something for me:
When you had
x² +2wx = v
How did you go from there to:
x² +2wx +w² -w² = v
I assume you added w² to both sides then wrapped x² +2wx +w² into the Binomial square: (x+w)²
leaving (x+w)² = w² + v
How come it isn't:
x² +2wx +w² = +w² +v
Which would then be
(x+w)² = w² +v
Where did the -w come after you had x² + 2wx = v?
Thanks
