What would you do if you were asked to find $x$ in the equation $x^2 = 45x + 23$? This is called a quadratic equation since it contains the unknown variable $x$ squared. The name name comes from the Latin quadratus, which means square. Quadratic equations come up very often so mathematicians came up with a general formula for solving these equations. We will learn about this formula in this section.

Before we can apply the formula, we need to rewrite the equation in the form \[ ax^2 + bx + c = 0, \] where we moved all the numbers and $x$s to one side and left only $0$ on the other side. This is the called the standard form of the quadratic equation. For example, to get the expression $x^2 = 45x + 23$ into the standard form, we can subtract $45x+23$ from both sides of the equation to obtain $x^2 - 45x - 23 = 0$. What are the values of $x$ that satisfy this formula?

The solutions to the equation \[ ax^2 + bx + c = 0, \] are \[ x_1 = \frac{-b + \sqrt{b^2-4ac} }{2a} \ \ \text{ and } \ \ x_2 = \frac{-b - \sqrt{b^2-4ac} }{2a}. \]

Let us now see how this formula is used to solve the equation $x^2 - 45x - 23 = 0$. Finding the two solutions is a simple mechanical task of identifying $a$, $b$ and $c$ and plugging these numbers into the formula: \[ x_1 = \frac{45 + \sqrt{45^2-4(1)(-23)} }{2} = 45.5054\ldots, \] \[ x_2 = \frac{45 - \sqrt{45^2-4(1)(-23)} }{2} = -0.5054\ldots. \]

This is an important proof. You should know how to derive the quadratic formula in case your younger brother asks you one day to derive the formula from first principles. To derive this formula, we will use the completing-the-square technique which we saw in the previous section. Don't bail out on me now, the proof is only two pages.

Starting from the equation $ax^2 + bx + c = 0$, our first step will be to move $c$ to the other side of the equation \[ ax^2 + bx = -c, \] and then to divide by $a$ on both sides \[ x^2 + \frac{b}{a}x = -\frac{c}{a}. \]

Now we must complete the square on the left-hand side, which is to say we ask the question: what are the values of $h$ and $k$ for this equation to hold \[ (x-h)^2 + k = x^2 + \frac{b}{a}x = -\frac{c}{a}? \] To find the values for $h$ and $k$, we will expand the left-hand side to obtain $(x-h)^2 + k= x^2 -2hx +h^2+k$. We can now identify $h$ by looking at the coefficients in front of $x$ on both sides of the equation. We have $-2h=\frac{b}{a}$ and hence $h=-\frac{b}{2a}$.

So what do we have so far: \[ \left(x + \frac{b}{2a} \right)^2 = \left(x + \frac{b}{2a} \right)\!\!\left(x + \frac{b}{2a} \right) = x^2 + \frac{b}{2a}x + x\frac{b}{2a} + \frac{b^2}{4a^2} = x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}. \] If we want to figure out what $k$ is, we just have to move that last term to the other side: \[ \left(x + \frac{b}{2a} \right)^2 - \frac{b^2}{4a^2} = x^2 + \frac{b}{a}x. \]

We can now continue with the proof where we left off \[ x^2 + \frac{b}{a}x = -\frac{c}{a}. \] We replace the left-hand side by the complete-the-square expression and obtain \[ \left(x + \frac{b}{2a} \right)^2 - \frac{b^2}{4a^2} = -\frac{c}{a}. \] From here on, we can use the standard procedure for solving equations. We put all the constants on the right-hand side \[ \left(x + \frac{b}{2a} \right)^2 = -\frac{c}{a} + \frac{b^2}{4a^2}. \] Next we take the square root of both sides. Since the square function maps both positive and negative numbers to the same value, this step will give us two solutions: \[ x + \frac{b}{2a} = \pm \sqrt{ -\frac{c}{a} + \frac{b^2}{4a^2} }. \] Let's take a moment to cleanup the mess on the right-hand side a bit: \[ \sqrt{ -\frac{c}{a} + \frac{b^2}{4a^2} } = \sqrt{ -\frac{(4a)c}{(4a)a} + \frac{b^2}{4a^2} } = \sqrt{ \frac{- 4ac + b^2}{4a^2} } = \frac{\sqrt{b^2 -4ac} }{ 2a }. \]

Thus we have: \[ x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 -4ac} }{ 2a }, \] which is just one step away from the final answer \[ x = \frac{-b}{2a} \pm \frac{\sqrt{b^2 -4ac} }{ 2a } = \frac{-b \pm \sqrt{b^2 -4ac} }{ 2a }. \] This completes the proof.

To have a proof we don't necessarily need to show the derivation of the formula as we did. The claim was that $x_1$ and $x_2$ are solutions. To prove the claim we could have simply plugged $x_1$ and $x_2$ into the quadratic equation and verified that we get zero. Verify on your own.

The golden ratio, usually denoted $\varphi=\frac{1+\sqrt{5}}{2}=1.6180339\ldots$ is a very important proportion in geometry, art, aesthetics, biology and mysticism. It comes about from the solution to the quadratic equation \[ x^2 -x -1 = 0. \]

Using the quadratic formula we get the two solutions: \[ x_1 = \frac{1+\sqrt{5}}{2} = \varphi, \qquad x_2 = \frac{1-\sqrt{5}}{2} = - \frac{1}{\varphi}. \]

You can learn more about the various contexts in which the golden ratio appears from the excellent wikipedia article on the subject. We will also see the golden ratio come up again several times in the remainder of the book.

Often times, we are interested in only one of the two solutions to the quadratic equation. It will usually be obvious from the context of the problem which of the two solutions should be kept and which should be discarded. For example, the time of flight of a ball thrown in the air from a height of $3$ meters with an initial velocity of $12$ meters per second is obtained by solving a quadratic equation $0=(-4.9)t^2+12t+3$. The two solutions of the quadratic equation are $t_1=-0.229$ and $t_2=2.678$. The first answer $t_1$ corresponds to a time in the past so must be rejected as invalid. The correct answer is $t_2$. The ball will hit the ground after $t=2.678$ seconds.

In the previous section we discussed the quadratic factoring operation by which we could rewrite a quadratic function as the product of two terms $f(x)=ax^2+bx+c=(x-x_1)(x-x_2)$. The two numbers $x_1$ and $x_2$ are called the roots of the function: this is where the function $f(x)$ touches the $x$ axis.

Using the quadratic equation you now have the ability to factor any quadratic equation. Just use the quadratic formula to find the two solutions $x_1$ and $x_2$ and then you can rewrite the expression as $(x-x_1)(x-x_2)$.

Some quadratic expression cannot be factored, however. These correspond to quadratic functions whose graphs do not touch the $x$ axis. They have no solutions (no roots). There is a quick test you can use to check if a quadratic function $f(x)=ax^2+bx+c$ has roots (touches or crosses the $x$ axis) or doesn't have roots (never touches the $x$ axis). If $b^2-4ac>0$ then the function $f$ has two roots. If $b^2-4ac=0$, the function has only one root. This corresponds to the special case when the function touches the $x$ axis only at one point. If $b^2-4ac<0$, the function has no real roots. If you try to use the formula for finding the solutions, you will fail because taking the square root of a negative number is not allowed. Think about it—how could you square a number and obtain a negative number?