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When I was in high school, they taught me to solve quadratic equations with this formula:

$$x=\frac{\sqrt{4 \text{ac}+b^2}-b}{2 a}$$

EDIT: The original formula is this one: $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$, I confused it with another formula on Wikipedia.

While reading Mathematics for the Nomathematician, Morris Kline suggests that a quadratic equation should be solved by making another equation with half of the coefficient of $b$:

$$y=x+\frac{b}{2}$$

Then:

$$x=y-\frac{b}{2}$$

and then proceed with the substitution on the quadratic equation, I guess I don't need to describe the rest as it may be a standard procedure.

Are there benefits on using the first formula? I also imagine that they could be related somehow, but I still can't see this relation.

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    Unless your quadratics are different from everyone else's, don't you want $-4ac+b^2$ under that square root?2012-08-03
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    Assuming $a=1$, Kline is describing "completing the square".2012-08-03
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    @Robert: Ah - this all suddenly makes sense to me.2012-08-03
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    @mixedmath Yes. I missed something.2012-08-03
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    @mixedmath This was also one of the suggetions on wikipedia, I'm just not sure if that's right.2012-08-03
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    About the $\sqrt{4ac+b^2}$: Conceivably they wrote the equation as $ax^2+bx=c$.2012-08-03
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    The benefit of using the quadratic formula is that you do not have to complete the square. The benefit of commpleting the square in a general context is to understand how the formula arises. The final benefit, hopefully, is the realization that abstraction (using a,b,and c in place of specific numbers) gives a general formula for the solution, relieving you of the burden of completing the square everytime you want to solve a quadratic. Unfortunately, this last benefit is hidden if the derivation is never discussed.2012-08-04
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    @GustavoBandeira: You might want to look at the [following link](http://math.stackexchange.com/questions/49229/why-can-all-quadratic-equations-be-solved-by-the-quadratic-formula) where I describe an algebraically smoother path to the Quadratic Formula.2012-08-11
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    @AndréNicolas You got a nice answer there. I'm kinda used to the way answering it by dividing all by a.2012-08-11

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It's the same thing. What Kline is doing is what's in general known as a depression, which is a variable substitution done for a polynomial of degree $n$, such that the resulting polynomial has no term of degree $x^{n-1}$.

One way to see how depression works is to consider the Vieta formulae, which in the case of the quadratic $ax^2+bx+c=a(x-x_1)(x-x_2)$ looks like this:

$$\begin{align*} x_1 x_2 &= \frac{c}{a}\\ x_1 + x_2 &= -\frac{b}{a} \end{align*}$$

From this, if the sum of the roots is $-\dfrac{b}{a}$, then the mean of the roots is $-\dfrac{b}{2a}$. Thus, the depression substitution

$$y=x+\frac{b}{2a}$$

can be geometrically interpreted as shifting the parabola corresponding to your quadratic $ax^2+bx+c$ such that it is "centered" about the origin, and the two roots are laid out symmetrically.

If we depress our original quadratic, we get

$$\require{cancel}\begin{align*} ax^2+bx+c&=a\left(y-\frac{b}{2a}\right)^2+b\left(y-\frac{b}{2a}\right)+c\\ &=a\left(y^2-\frac{b}{a}y+\frac{b^2}{4a^2}\right)+b\left(y-\frac{b}{2a}\right)+c\\ &=ay^2\cancel{-by}+\frac{b^2}{4a}\cancel{+by}-\frac{b^2}{2a}+c\\ &=ay^2+\frac{b^2}{4a}-\frac{2b^2}{4a}+c\\ &=ay^2-\frac{b^2}{4a}+c=ay^2-\frac{b^2-4ac}{4a} \end{align*}$$

and I'm sure you know how easy it is to solve the equation

$$ay^2-\frac{b^2-4ac}{4a}=0$$

Having solved for $y$, you undo the depression you did, which means you have to add the term $-\dfrac{b}{2a}$ to get the actual roots you want. That's where that part of the quadratic formula comes from.

In short, I would not say that Kline's method is the most expedient, but it at least looks to me that this slower method allows for more cogitation on what the steps are supposed to "mean", as opposed to a lazy plug-and-chug.

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    This is ... great!2012-08-04
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    I understand it (the process of solving a quadratic equation) mechanicaly, I can solve a quadratic equation but the procedure you explained is still a little obscure to me: It seems I can understand a little of this heuristic but no very deeply. Can you suggest some reading about it?2012-08-04
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    I've also googled a little about depresion(Using "depresion mathematics" as search subject), and the only thing I've found was about people getting depressed while studying/doing mathematics - The search couldn't be deeper because of my slow internet.2012-08-04
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    @Gustavo, the search term you should be using is `polynomial depression`; you'll see that the general idea is also used in deriving explicit solutions for the cubic and quartic equations.2012-08-04
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    @J.M. Only in cubic and quartic? Can't it solve polynomials on any degree?2012-08-04
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    @Gustavo, depression can be done on *any* polynomial, but you should know that the *general* formulae for expressing the roots of polynomials of degree five and higher require "special functions"; Abel and others have shown that there are no general formulae for those roots that only involve elementary functions.2012-08-04
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    In the same book, I don't understand why he works with $x^2+px+q=0$ instead of $ax^2+bx+c=0$.2012-08-04
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    Well, things are slightly simpler if the polynomial is *monic* (that is, the leading coefficient is $1$), and when solving a quadratic, one can always perform a division so that you are working with a monic polynomial.2012-08-04
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    I can't see the division in that polynomial, considering it's $ax^2+bx+c=0$, shouldn't it be $x^2+\frac{bx}{a}+\frac{c}{a}=0$?2012-08-04
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    @Gustavo, congratulations; you've just formed a monic quadratic equation from the general form.2012-08-04
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    Then from here, I proceed with the same procedure: $y=x+\frac{\frac{bx}{a}}{2}$?2012-08-04
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    There should not be an $x$ on the right-hand side of your expression.2012-08-04
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    Yes. I'm just avoiding to write the rest of the procedure, I know that I should make: $x=y-\frac{\frac{bx}{a}}{2}$.2012-08-04
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    You did not understand me; you wrote $y=x+\frac{bx}{2a}$, which is wrong. The correct depression relation is $y=x+\frac{b}{2a}$, which, if you read my answer, I have mentioned very much earlier.2012-08-04
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    I can't understand why the $x$ disappears from the $bx$.2012-08-04
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    @Gustavo, why not try it yourself? Compare what happens if you retain the $x$, versus if you do what I said to do.2012-08-04
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    Yep. I'll try it, I'll also google a little for polynomial depression. Thanks for the help.2012-08-04
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    @J.M. I finally discovered it! The procedure says: Half of the coefficient of $x$ which is $\frac{b}{2a}$ and not $\frac{bx}{2a}$. It's the second time I make the same mistake. I hope not making it again.2012-08-06
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    @Gustavo, finally. The coefficient of $x$ is $\frac{b}{a}$, not $\frac{bx}{a}$. The $x$ is not supposed to be there; it's the variable. The coefficient is *everything else* multiplying the variable. Clear?2012-08-06
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    Yep. I'm happy now. =)2012-08-06
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    I'm also happy with something I just read: When doing mathematics, Stubbornness is more important than inteligence. It's cool because it's the only thing I have. Haha2012-08-06
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    Oh, I almost forgot: Thanks for your time.2012-08-06
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Let's try to follow Kline's advice (and all the time we assume the coefficients $\,a,b,c\,$ are real).

We have $\,ax^2+bx+c=0\,\,,\,a\neq 0\,$ and we make the substitution $$y:=x+\frac{b}{2}\Longleftrightarrow x = y-\frac{b}{2}$$ and from here, substituting in the equation

$$a\left(y-\frac{b}{2}\right)^2+b\left(y-\frac{b}{2}\right)+c=0$$ ...and now? But for messing up the equation's variable I can't see any real advance or simplification.

I think what he actually meant, or should have meant, is the following, which is only the rather well-known method of completing the square (CS): $$X^2\pm BX=\left(X \pm \frac{B}{2}\right)^2-\frac{B^2}{4}\,\,,\,\text{and from here}$$ $$0=ax^2+bx+c\,\Longrightarrow\,x^2+\frac{bx}{a}=-\frac{c}{a}\stackrel{CS}\Longrightarrow\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}=-\frac{c}{a}\,\Longrightarrow$$ $$\Longrightarrow\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\,\,\,\text{(see the numerator?!)}\,\,\stackrel{\text{sq.rt. in both sides}}\Longrightarrow$$ $$\Longrightarrow x_{1,2}+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}\,\Longrightarrow\,x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ .

So, as you can see, the benefits is that you need that expression to solve quadratics..as simple as that!

Of course, the number $\,\Delta:=b^2-4ac\,$ , called the quadratic's discriminant serves to know beforehand about the above equation's possible solutions: $$\begin{align*}\Delta>0\Longrightarrow & \,\text{there exist two different real solutions to the equation}\\ \Delta=0\Longrightarrow & \,\text{there exists only one unique real solution to the equation}\\ \Delta<0\Longrightarrow & \,\text{the equation has no real solutions}\end{align*}$$

In the last case above there exist two conjugate complex non-real solutions (disregard this if you haven't yet studied complex numbers)

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    What's the meaning of $:=$? It's the first time I see it.2012-08-04
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    := means the left side is defined by the right one2012-08-11
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    "there exists only one unique real solution to the equation" - slightly strange wording, like, there are actually two solutions, but the other one isn't unique? :)2012-08-11
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    @BenMillwood, I think that's a language fix. Where I live (Israel), even at 7th-8th grade in junion high school, children are taught to say "there are (there aren't) *REAL* solution(s) to this quadratic equation", apparently as preparation for complex numbers later in high school. The meaning, though, is correct, I believe, under the proper interpretation: one *unique real* solution means there's only one number that solves the quadratic eq.2012-08-11
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    @DonAntonio: I see where you're coming from, but I think it's strange to use "unique" here - what are you saying that isn't said by "there exists only one real solution to the equation"? If you think the latter is untrue because there is a repeated solution, then you think there are two solutions, but then you say that there is one unique solution, which just leads me to wonder if the other one is unique, too...2012-08-11
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    I think my difficulty is that there's an implication that it's possible to have more than one unique solution, which is kind of a contradiction in terms.2012-08-11