We have previously classified polynomials by the number of terms they contain (e.g., monomials, binomials, trinomials, etc.) However, we can classify polynomials -- and equations involving them -- by their degree as well.

The last section introduced linear functions, which are defined by a polynomial of degree $1$ -- namely, something of the form $mx+b$. In a similar way, we define a polynomial of degree $2$ as a **quadratic polynomial** (from the Latin word *quadraticus*, meaning "*made square*"). Naturally, equations involving these, often written in the form $ax^2 + bx + c = 0$ for real values $a$, $b$, and $c$, are called **quadratic equations**.

So far, all of the quadratic equations we have solved have involved quadratic polynomials that either only have a single occurrence of the variable in question, or those that easily factor by inspection. However, we can solve other quadratic equations as well. This section explores a number of methods for solving quadratic equations, including the strategy known as "completing the square", a new modern method developed by Po-Shen Loh, the older "method of depressed terms", and -- of course -- the famed quadratic equation.

The completing the square technique is useful beyond just solving quadratic equations -- particularly in calculus when one must "massage" and expression to fit a certain form before continuing to do something else to it. As such, it is important to be aware of this strategy even when one has other methods to solve a general quadratic equation.

The general idea behind completing the square in solving quadratic equations is to manufacture a scenario in which "socks and shoes" can be applied. Consequently, we seek to rewrite something of this form: $ax^2 + bx + c = 0$ (which has $2$ occurrences of the variable) as an equation where we see only $1$ occurrence of the variable.

We know that if the two occurrences of the variable (say, $x$ here) were found in an expression of the form $x^2 + 2bx + b^2$ for some $b$, then we could rewrite that expression as the square of a binomial $(x+b)^2$, thereby reducing the two occurrences of the variable down to a single occurrence.

For example, suppose you were solving $$x^2 + 6x + 9 = 7$$ Rewriting this as $(x+3)^2 = 7$, we can then solve from there using "socks and shoes". (i.e., Take the square root, "plus-or-minus", of both sides to get $x+3 = \pm\sqrt{7}$. Then subtract $3$ from both sides to find the solution $x = -3 \pm \sqrt{7}$).

However, what if the left side of the original equation wasn't so conveniently given as a square of a binomial? For example, consider the following problem:

We see here that the left side is (sadly) Recall the form for the square of a binomial is $(x+b)^2 = x^2 + 2bx + b^2$. So if we equate $x^2 + 14x$ with $x^2 + 2bx$, we can see that we must have $b=7$ here. Note, the value of $b$ we seek will always be half of the coefficient on the linear (i.e., degree $1$) term. As such, we would love to see $x^2 + 14x + 49$ (noting $b^2 = 49$) on the left side of our equation instead of the $x^2 + 14x + 3$ we currently have. So let us add $49$ to the left hand side! Of course, we can't just do that "willy-nilly" -- that would change the equation! However if we added $49$ to the left side and then subtracted it right off again (i.e., adding a well-chosen value of zero to the left side), the resulting equation stays equivalent to the first, and we can write $x^2 + 14x + 49$ as $(x+7)^2$ which allows us to complete solving the equation with the "socks and shoes" principle:$\require{color}$ $$\begin{array}{rcl} x^2 + 14x + 3 &=& -2\\ x^2 + 14x {\color{red} {} + 49 - 49} + 3 &=& -2\\ (x^2 + 14x + 49) - 49 + 3 &=& -2\\ (x+7)^2 - 49 + 3 &=& -2\\ (x+7)^2 &=& 44\\ x + 7 &=& \pm \sqrt{44} = \pm 2\sqrt{11}\\ x &=& -7 \pm 2\sqrt{11} \end{array}$$ |

If the quadratic term (i.e., the term involving the square) has a coefficient other than one, we can simply factor out this coefficient from the first two terms. However, we must be particularly careful regarding any well-chosen value of zero we choose to add to one side of our equation -- as the next example demonstrates:

Again, we don't see a perfect square on the left, so we must manufacture one. First, factor out the $3$ from the first two terms on the left to make this easier to do: $$3(x^2 - \textstyle{\frac{5}{3}}x) + 1 = 0$$ Then, we again realize that to make the first two terms inside the parentheses $(x^2 - \frac{5}{3})$ match the first two terms of the square of some binomial $(x + b)^2 = x^2 + 2bx + b^2$, we need $b$ equal to half the coefficient on the linear term (i.e., $b = -\frac{5}{6}$) We carefully add on the We then carefully replace $x^2 - \frac{5}{3} x + \frac{25}{36}$ with the corresponding binomial square to find $$\textstyle{3((x-\frac{5}{6})^2 -\frac{25}{36}) + 1 = 0}$$ Now, we proceed with socks and shoes (noting in particular that the $-\frac{25}{36}$ is |

In mathematics, discovering a new solution to an old problem can be almost as exciting discovering the first solution to an unsolved problem. While geometric methods for solving certain quadratic equations existed as far back as to be shown on cuneiform tablets from ancient Babylonia, and rules for solving quadratic equations appear in the Chinese

Amazingly, in September of 2019 a *new* approach to solving quadratic equations has been found -- although its creater, Carnegie Melon professor Dr. Po-Shen Loh, would argue the seeds for his idea can be traced all the way back to the aforementioned Babylonians.

The following details the mechanics of his approach in a couple of different cases:

First we consider an example where the related quadratic polynomial "factors nicely":

One often hopes that the quadratic expression in a quadratic equation will "factor nicely" so that we can solve the equation quickly via the zero-product property. However, very often students will attempt to factor the quadratic expression in question thorugh a series of "guess-and-checks". This is typically quick for equations $ax^2 + bx + c = 0$ when $a$ and $c$ don't have many factors to consider, but can get tedius otherwise. Dr. Loh's method takes the guesswork out of the picture. Let us consider only quadratics where $a=1$, as we could always divide both sides by something appropriate to force this to happen. Then of course, we hope that the resulting quadratic expression (like the $x^2 -2x - 24$ in this problem) factors into something nice like: $$(x - r_1)(x - r_2)$$ which when expanded equals $$x^2 - (r_1 + r_2) x + (r_1 r_2)$$ In this particular case, the expanded form is $x^2 - 2x - 24$, so we immediately see that we seek $r_1$ and $r_2$ where $r_1 + r_2 = 2$ and $r_1 r_2 = -24$. Dr. Loh makes the following observation: the average value of $r_1$ and $r_2$ is $\cfrac{r_1 + r_2}{2}$, and we know the value of $r_1 + r_2$! For this problem, since $r_1 + r_2 = 2$, the average value of $r_1$ and $r_2$ must be $1$. Now note that the average of any two numbers is the value at the midpoint between them on the number line. As such, let us call the two values we seek $(1-u)$ and $(1+u)$. This allows us to make the following sequence of deductions (mostly by "socks and shoes"): $$\begin{array}{rcl} (1+u)(1-u) &=& -24\\ 1-u^2 &=& -24\\ u^2 &=& 25\\ u &=& 5 \quad {\scriptstyle \textrm{(note, interpreting $u$ as a distance we only keep the positive value)}} \end{array}$$ Thus, $$r_1 = 1-5 = -4 \quad \textrm{ and } \quad r_2 = 1+5 = 6$$ and consequently $x^2 - 2x - 24 = 0$ reduces to $(x+4)(x-6) = 0$, with both having solutions $$x=-4 \quad \textrm{ and } \quad x=6$$ |

Now let's consider an example when the factorization is not so "pretty":

We need a leading coefficient of $1$ on the related quadratic to proceed as we did above, so let us divide both sides by $3$ and solve for the zeros of the following instead: $$x^2 + \textstyle{\frac{7}{3}x + \frac{1}{3}}$$ We again equate this to $(x-r_1)(x-r_2) = x^2 - (r_1 + r_2)x + (r_1 r_2)$ to determine $r_1 + r_2 = -\frac{7}{3}$ and $r_1 r_2 = \frac{1}{3}$. Consequently, the average of $r_1$ and $r_2$ is $\frac{r_1 + r_2}{2} = -\frac{7}{6}$. So let $r_1 = -\frac{7}{6} + u$ and $r_2 = -\frac{7}{6} - u$, given this average should be value precisely in the middle of $r_1$ and $r_2$. Given $r_1 r_2 = 2$, we then have: $$\begin{array}{rcl} (-\frac{7}{6} + u)(-\frac{7}{6} - u) &=& \frac{1}{3}\\ \frac{49}{36} - u^2 &=& \frac{1}{3}\\ u^2 &=& \frac{37}{36}\\ u &=& \frac{\sqrt{37}}{6} \quad {\scriptstyle \textrm{ again interpreting $u$ as a distance, which is then positive}}\\ \end{array}$$ Thus, $$r_1 = \textstyle{-\frac{7 - \sqrt{37}}{6} \quad \textrm{ and } \quad r_2 = -\frac{7 + \sqrt{37}}{6}}$$ and consequently $3x^2 + 7x + 1 = 0$ reduces to $3(x+\frac{7-\sqrt{37}}{6})(x+\frac{7+\sqrt{37}}{6}) = 0$, with both having solutions $$x = \frac{-7 \pm \sqrt{37}}{6}$$ |

The third method of solving quadratic equations explored below not only helps us with this task, but also leads to a technique that can be used to solve a general cubic equation (i.e., an equation of the form $ax^3 + bx^2 + cx + d = 0$).

The strategy employed in this case is to horizontally shift $f(x) = a_f x^2 + b_f x + c_f$ by some amount $h$ to produce a related function $g(x) = f(x+h)$ that has aEssentially, we want to find the value $h$, so that $g(x) = a_g x^2 + c_g$. Solutions to $g(x) = 0$ can then be found using "socks and shoes", and these can then be easily transformed into solutions to $f(x) = 0$ by simply "shifting them back".

The shift $h$ that we need is always of the same form, so let us find it, before working any specific examples. Note that $$\begin{array}{rcl} g(x) &=& f(x+h)\\ &=& a_f (x+h)^2 + b_f (x+h) + c_f\\ &=& a_f (x^2 + 2xh + h^2) + b_f (x+h) + c_f\\ &=& a_f x^2 + 2a_f h x + a_f h^2 + b_f x + b_f h + c_f\\ &=& (a_f) x^2 + (2a_f h + b_f) x + (a_f h^2 + b_f h + c_f) \end{array}$$ If we wish no linear term, we'll need the value of $h$ that ensures the coefficient on $x$ above is zero.

As such, we solve for the $h$ we need, and then notice the constant term of $g(x)$ just happens to be $f(h)$.

$$2 a_f h + b_f = 0 \quad \longrightarrow \quad h = \cfrac{-b_f}{2 a_f}, \quad g(x) = a_f x^2 + f(h)$$ Now, let us consider how we can use these two facts to solve a quadratic equation:
Note that $h = \frac{12}{(2)(2)} = 3$, and since $f(3) = 18 - 36 + 9 = -9$, we must have $g(x) = 2x^2 -9$ Solving $g(x) = 0$, we have $$\begin{array}{rcl} 2x^2 - 9 &=& 0\\ 2x^2 &=& 9\\ x^2 &=& \frac{9}{2}\\ x &=& \pm \frac{3}{\sqrt{2}} \end{array}$$ Now, we simply "shift back" by noticing if $x$ is a solution to $g(x) = 0$, then it must also be a solution to $f(x+h) = 0$. Consequently, $x + h$ will be a solution to $f(x) = 0$. Here, that means $f(x) = 0$ is solved by $$\pm \frac{3}{\sqrt{2}} + 3$$ While not necessary -- commuting the terms, rationalizing the denominator, and combining things into a single fraction reveals these solutions are identical in value but just different in form from the solutions naturally produced by either of the methods first discussed: $$x = 3 \pm \frac{3\sqrt{2}}{2} = \frac{6 \pm 3\sqrt{2}}{2}$$ |

Any of the above methods can be applied to a general quadratic of the form $ax^2 + bx + c = 0$ to produce a formula in terms of $a$, $b$, and $c$ that yields this equation's solutions. Below, we use the method of depressed terms to this end:

Note that for $f(x) = ax^2 + bx + c$, we again have $h = \frac{-b}{2a}$.

Then observe that $$\begin{array}{rcl} \displaystyle{f\left(\frac{-b}{2a}\right)} &=& \displaystyle{a \left(\frac{-b}{2a}\right)^2 + b\left(\frac{-b}{2a}\right) + c}\\\\ &=& \displaystyle{\frac{b^2}{4a} - \frac{b^2}{2a} + c}\\\\ &=& \displaystyle{\frac{-b^2}{4a} + c\frac{4a}{4a}}\\\\ &=& \displaystyle{\frac{-b^2 + 4ac}{4a}} \end{array}$$ Now we find solutions to $g(x) = 0$, recalling that $g(x) = ax^2 + f(\frac{-b}{2a})$: $$\begin{array}{rcl} ax^2 + f(\frac{-b}{2a}) &=& 0\\\\ ax^2 &=& -f(\frac{-b}{2a})\\\\ x^2 &=& \displaystyle{\frac{-f(\frac{-b}{2a})}{a}}\\\\ x^2 &=& \displaystyle{\frac{b^2 - 4ac}{4a^2}}\\\\ x &=& \displaystyle{\pm \sqrt{\frac{b^2 - 4ac}{4a^2}}}\\\\ x &=& \displaystyle{\pm \frac{\sqrt{b^2 - 4ac}}{2a}} \end{array}$$ Finally, we add $\displaystyle{h=\frac{-b}{2a}}$ to this solution to $g(x) = 0$ to find a solution to $f(x) = 0$: $$x = \frac{-b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$ Thus, upon combining the terms into a single fraction, we establish the following:

The solutions to $ax^2 + bx + c = 0$ are given by $$\displaystyle{x =\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$$ |

Returning to Po-Shen Loh's method for a moment, notice that one fact on which this method relies is that products of the form $(x-r_1)(x-r_2)$, when expanded reveal coefficients that have a beautiful symmetry to them: $$x^2 - (r_1 + r_2)x + (r_1 r_2)$$ Importantly, note that if we swap the positions of $r_1$ and $r_2$ above, the coefficients stay the same. In this way, we say that the coefficients of a quadratic (with leading coefficient $1$) are each a symmetric function of its roots involving only sums and products of those roots.

One might naturally wonder if the same can be said for higher degree polynomials (again with leading coefficient $1$) -- and one can!

Consider the expansion of $(x-r_1)(x-r_2)(x-r_3)$ into $x^3 + bx^2 + cx + d$, where we get:
$$\begin{array}{rcl}
b &=& -(r_1 + r_2 + r_3)\\
c &=& +(r_1 r_2 + r_1 r_3 + r_2 r_3)\\
d &=& -(r_1 r_2 r_3)
\end{array}$$
and the expansion of $(x-r_1)(x-r_2)(x-r_3)(x-r_4)$ into $x^4 + bx^3 + cx^2 + dx + e$, where we get:
$$\begin{array}{rcl}
b &=& -(r_1 + r_2 + r_3 + r_4)\\
c &=& +(r_1 r_2 + r_1 r_3 + r_1 r_4 + r_2 r_3 + r_2 r_4 + r_3 r_4)\\
d &=& -(r_1 r_2 r_3 + r_1 r_2 r_4 + r_1 r_3 r_4 + r_2 r_3 r_4)\\
e &=& +(r_1r_2 r_3 r_4)
\end{array}$$
In both cases above, if you *permute*^{†} the placements of $r_1$, $r_2$, $r_3$, and $r_4$ above in any way you like, the values of the coefficients never change.

The general result that governs this symmetry for all polynomials is known as Viete's Formulas, named after 16th-century French mathematician Francois Viete, and will play a prominent role in our later proof of Abel's Impossibility Theorem.

*†
This perhaps feels different than the "swapping" to which we earlier referred -- but recall that swapping two things is no different than transposing them, and transpositions are just the special case of a permutation involving only two elements.*

That said, one can use Viete's formulas in other ways as well. They can be a particularly effective way to find the values of other symmetric expressions involving the roots of a polynomial function -- as the next example demonstrates:

That was quick, wasn't it! |