Think of the tune to \u201cPop Goes the Weasel.\u201d Now sing these lyrics:<\/p>\n

Neg-a-tive b<\/em>, plus or minus
The square root of b<\/em> squared
mi-nus four a<\/em> c<\/em>
All! over two a<\/em><\/p>\n

This jingle has helped generations of algebra students recall the quadratic formula that solves every equation of the form $latex ax^2+bx+c=0$. The formula is as useful as it is likely to appear in the dictionary under \u201cmath anxiety,\u201d and a quick look shows you why:<\/p>\n

$$frac{-bpm sqrt{b^2-4ac}}{2a}$$<\/p>\n

As intimidating as this looks, hiding inside is a simple secret that makes solving every quadratic equation easy: symmetry. Let\u2019s look at how symmetry makes the quadratic formula work and how a lack of symmetry makes solving cubic equations (of the form $latex ax^3+bx^2+cx+d =0$)\u00a0much, much harder. So much harder, in fact, that a few mathematicians in the 1500s spent their lives embroiled in bitter public feuds competing to do for cubics what was so easily done for quadratics.<\/p>\n

Solving equations is a core skill in math class \u2014 it helps us find maximum profits, minimum distances, points of intersection, and much more. One of the most basic equations we learn to solve is $latex f(x)=0$. Given a function $latex f(x)$, this equation asks: What inputs x <\/em>return an output of 0? For this reason, solutions to this equation are sometimes called the \u201czeros\u201d or \u201croots\u201d of the function.<\/p>\n

Before we find the roots of every quadratic function, let\u2019s start with an easy one: What are the roots of $latex f(x)=x^2-9$? To find them, just solve the equation $latex f(x)=0$.<\/p>\n

$latex f(x)=0$
$latex x^2-9=0$
$latex x^2=9$
$latex x=pm3$<\/p>\n

These roots are easy to find because this equation is easy to solve. All you have to do is isolate x<\/em>. Notice that we need that $latex\u00a0pm$ in the last line, because both 3 and -3 have the property that when you square them you get 9. A quick check that $latex f(3)=f(-3)=0$ verifies that these are indeed the inputs that make $latex f(x)$ output 0.<\/p>\n

That $latex\u00a0pm$ also points to the symmetry inherent in the situation. The quadratic function has two roots, and if you imagine the two roots on a number line, you\u2019ll see that they are symmetric about $latex x=0$.<\/p>\n

And when you remember that the graph of a quadratic function is a parabola, this makes a lot of sense. Every parabola has an axis of symmetry that splits the parabola into two mirror-image pieces. In the case of $latex f(x)=x^2-9$, the axis of symmetry is the y<\/em>-axis (the line $latex x=0$). When you graph $latex f(x)=x^2-9$ in the usual way, by treating\u00a0x\u00a0<\/i>as the independent variable and setting $latex y=f(x)$, you can see its roots on the\u00a0x<\/i>-axis, equidistant from and on either side of the\u00a0y<\/i>-axis.<\/p>\n

For a more complicated quadratic like $latex f(x)=x^2-8x-9$,\u00a0finding the roots takes a bit more digging.<\/p>\n

$latex f(x)=0$
$latex x^2-8x-9=0$
$latex x^2-8x=9$<\/p>\n

We can set $latex f(x)$ equal to 0 and move the 9 to the right side as we did before,\u00a0but we can\u2019t take the square root of both sides to isolate x<\/em>. That other term with the x<\/em> in it is standing in the way. But this function, like every quadratic, is symmetric, and we can use that symmetry to navigate around the problem. We just need a little algebra to make the symmetry more transparent.<\/p>\n

Let\u2019s rewrite the function $latex f(x)=x^2-8x-9$ as $latex f(x)=x(x-8)-9$. Now focus on the $latex x(x-8)$ part. This will be equal to 0 in two situations \u2014\u00a0if x = <\/em>0 or if x = <\/em>8 \u2014\u00a0and this guarantees that $latex f(0)$ and $latex f(8)$ will take the same value of -9. This gives us two symmetric points on the parabola, and since the axis of symmetry has to split $latex x=0$ and $latex x=8$ down the middle, it must be the line $latex x=4$.<\/p>\n

Now that we\u2019ve found the symmetry, it\u2019s time to leverage it. We\u2019re going to shift our parabola four units to the left so that its axis of symmetry moves from the line $latex x=4$ to the line $latex x=0$. There\u2019s a simple way to perform this translation algebraically: We replace every x<\/em> with x <\/em>+ 4.<\/p>\n

Let\u2019s call $latex g(x)$ the new quadratic function we get when we replace x<\/em> with x<\/em>+ 4. In other words, let $latex g(x)=f(x+4)$. Watch what happens when we simplify $latex g(x)$:<\/p>\n

$latex g(x)=f(x+4)$
$latex g(x)=(x+4)^2-8(x+4)-9$
$latex g(x)=x^2+8x+16-8x-32-9$
$latex g(x)=x^2-25$<\/p>\n

After we apply the distributive property a few times and collect like terms, the\u00a0x\u00a0<\/i>term of our new translated quadratic vanishes, and this makes finding the roots of $latex g(x)$ easy:<\/p>\n

$latex g(x)=0$
$latex x^2-25=0$
$latex x^2=25$
$latex x=pm5$<\/p>\n

The roots of $latex g(x)$ are $latex x=pm5$, so to find the roots of $latex f(x)=x^2-8x-9$, we just move the roots of $latex g(x)$ back four units to the right. This us gives us the roots of $latex f(x)$: $latex 4pm5$, or 9 and -1, which you can verify by computing $latex f(9)=f(-1)=0$.<\/p>\n

The secret to solving this slightly harder quadratic equation was to slide it over and turn it into an easier quadratic equation by eliminating the interfering\u00a0x\u00a0<\/i>term. This approach will work on any quadratic function. Given an arbitrary quadratic $latex f(x)=ax^2+bx+c$, \u00a0you can always find its axis of symmetry with the same bit of factoring:<\/p>\n

$latex f(x)=ax^2+bx+c$
$latex f(x)=x(ax+b)+c$<\/p>\n

In this form you can see that $latex f(0)=fleft(-frac{b}{a}right)=c$, which means the axis of symmetry is halfway between $latex x=0$ and $latex x=-frac{b}{a}$. In other words, the axis of symmetry of any quadratic function $latex f(x)=ax^2+bx+c$ is the line $latex x=-frac{b}{2a}$.\u00a0And this should look familiar. It\u2019s hiding in the quadratic formula!<\/p>\n

$$ x=frac{-bpmsqrt{b^2-4ac}}{2a}$$<\/p>\n

It\u2019s easier to see if you rewrite it like this:<\/p>\n

$$ x=-frac{b}{2a}pmfrac{sqrt{b^2-4ac}}{2a}$$<\/p>\n

The quadratic formula relies on the fact that the roots of the quadratic $latex f(x)=ax^2+bx+c$ are symmetric about $latex x=-frac{b}{2a}$. And just as we did above, you can use that symmetry to find them: Just translate $latex f(x)$ by $latex -frac{b}{2a}$. This has the effect of eliminating the\u00a0x\u00a0<\/i>term, which allows you to then easily isolate\u00a0x <\/i>and solve. Do this, and you\u2019ll get the quadratic formula. (See the exercises below for more details.) This isn\u2019t as easy as humming a children\u2019s tune, but it demonstrates the important algebraic and geometric connections that make this formula work.<\/p>\n

Solving quadratics with the power of symmetry might embolden us to try a similar tactic on cubic equations. But while cubics do have symmetry, it\u2019s not the kind that helps with solving equations like $latex f(x)=0$.\u00a0Cubic graphs have \u201cpoint symmetry,\u201d which means there\u2019s a special point on the graph of every cubic function where, if a line passes through that point and intersects the cubic anywhere else, it intersects the graph again symmetrically about that point.<\/p>\n

This is a strong type of symmetry, but it doesn\u2019t help with finding roots. That\u2019s because the roots of a function occur where its graph crosses the horizontal line $latex y=0$ (the x<\/em>-axis),\u00a0and in general, those intersections aren\u2019t symmetric about the cubic\u2019s special point of symmetry.<\/p>\n

In fact, a cubic might have only root. No symmetry there.<\/p>\n

Yet, there\u2019s something from our earlier work with quadratics that can help.<\/p>\n

If we have a quadratic function $latex f(x)=ax^2+bx+c$ and we know its roots are $latex r_1 $ and $latex r_2$, then we can always write $latex f(x)$ in \u201cfactored\u201d form: $latex f(x)=a(x-r_1)(x-r_2)$. Now, when we multiply this out and simplify, we get something very useful to work with.<\/p>\n

$latex f(x)=a(x-r_1)(x-r_2)$
$latex f(x)=a(x^2-xr_2-r_1x+r_1r_2)$
$latex f(x)=a(x^2-(r_1+r_2)x+r_1r_2)$
$latex f(x)=ax^2-a(r_1+r_2)x+ar_1r_2$<\/p>\n