What are the benefits of using $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ to solve quadratic equations?

Question

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

$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}.$$

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.

Answer 1

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.

Answer 2

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)

Answer 3

The given answers are good, but let me show you the "typical" way of proving the original formula. The goal is to find the roots of the quadratic, so that:

$$Ax^2 + Bx + C = 0$$

The $C$ constant is a little messy, so let's move it to the other side:

$$Ax^2 + Bx = -C $$

Now, on the left-hand side, we can "complete the square", i.e., find out a value that will allow us to write the left-hand side as $(\_ + \_)^2$. This is easiest to do if the $x^2$ is by itself. So we'll divide both sides by $A$ before continuing.

$$x^2 + \frac{B}{A}x = -\frac{C}{A}$$

Now, when completing the square, the middle term is key. The middle term is twice the left-hand times the right-hand side. Here, since we only have $x^2$ on the left, the left-hand side of our square will obviously be $x$. The middle term is $\frac{B}{A}x$. The $x$ obviously comes from the left-hand side of our square. $\frac{B}{A}$ comes from the right side, but, remember, the middle term gets doubled in squares. Therefore, the right-hand side of the square should be $\frac{B}{2A}$, which means that the right-hand term will be the square of that, or $\frac{B^2}{4A^2}$. If we add this term to both sides we get:

$$ x^2 + \frac{B}{A}x + \frac{B^2}{4A^2} = -\frac{C}{A} + \frac{B^2}{4A^2} $$

The left-hand side is now a square. So we can simplify it to:

$$ \left(x + \frac{B}{2A}\right)^2 = -\frac{C}{A} + \frac{B^2}{4A^2}$$

Square-rooting both sides gets:

$$ x + \frac{B}{2A} = \pm \sqrt{-\frac{C}{A} + \frac{B^2}{4A^2}} $$

Subtracting $\frac{B}{2A}$ from both sides gets:

$$ x = \pm \sqrt{-\frac{C}{A} + \frac{B^2}{4A^2}} - \frac{B}{2A} $$

This is a quadratic formula, but not in the form we are used to seeing. However, it is just a matter of rearranging to get the quadratic formula.

First, let's simplify the interior of the square root. We will start by putting everything over the same denominator ($4A^2$).

$$ x = \pm \sqrt{\frac{-4AC + B^2}{4A^2}} - \frac{B}{2A} $$

The bottom part of the fraction is actually a square, so it can be pulled out of the square root.

$$ x = \frac{1}{2A} \pm \sqrt{-4AC + B^2} - \frac{B}{2A} \\ x = \frac{\pm \sqrt{-4AC + B^2}}{2A} - \frac{B}{2A} \\ x = \frac{\pm \sqrt{-4AC + B^2} - B}{2A} $$

This is the formula, just in the wrong order. Changing the order (both inside and outside the square root) yields:

$$ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} $$