A quadratic is a function, polynomial, equation, expression... with a power of $2$ in it.
The simple quadratic form follows the notation $f(x) = ax^2 + bx + c$.
Solving quadratics is really just a special case of the more general task of finding the zeros of a function, when that function is a second-degree polynomial (an expression where some term is raised to the power of two).
Finding the zeros of a function means finding the value(s) of $x$ such that $f(x)=0$. Those *zeros* (values of $x$) correspond to the $x$-intercepts of the function on its graph.
A quadratic function can be visualized with a graph like below, with [[Graphing Functions#Transformations|transformations]] applied according to the values of $a$, $b$ and $c$.
Depending on the values of the coefficients $a$, $b$ and $c$ :
- The graph may have two distinct $x$-intercepts (i.e. two real roots)
- The solution may be a double root (the same solution twice, like $\pm\sqrt{0}$), meaning the graph *touches* the $x$-axis at exactly one point, as shown in the above representation
- The equation may have no real solution, which occurs when solving outputs a complex or imaginary number (such as $\sqrt{-1}$, which is also written $i\sqrt{1}$)
To start solving a quadratic function and find its $x$-intercepts, we set the function equal to zero, since, again, we're looking for the zeros, if any, of the function.
$
\begin{align}
f(x) = 0
\\
ax^2 + bx + c = 0
\end{align}
$
There are a few techniques we can rely on to solve quadratics, which we'll explore in reverse order of complexity.
## Square Root Method
### $f(x)=x^2-18$
$
\begin{align}
x^2-18=0
\\
x^2=18
\\
\sqrt{x^2}=\sqrt{18}
\\
x=\pm\sqrt{9 \cdot 2}
\\
x=\pm3\sqrt{2}
\end{align}
$
We see that there are two real solutions, hence the graph of this quadratic function has two $x$-intercepts at $(-3\sqrt{2}, 0)$ and $(3\sqrt{2}, 0)$.
### $g(x)=3x^2-33$
$
\begin{align}
3x^2-33 = 0
\\
3x^2=33
\\
\frac{3x^2}{3}=\frac{33}{3}
\\
x^2=11
\\
x=\pm\sqrt{11}
\end{align}
$
Again, we find two real solutions. The $x$-intercepts of the function are $(-\sqrt{11},0)$ and $(\sqrt{11},0)$.
### $h(x)=(3x-2)^2+75$
$
\begin{align}
(3x-2)^2+75=0
\\
(3x-2)^2=-75
\\
\sqrt{(3x-2)^2}=\sqrt{-75}
\\
3x-2=\sqrt{25\cdot-3}
\\
3x-2=5i\sqrt{3}
\\
3x=5i\sqrt{3}+2
\\
x=\frac{5i\sqrt{3}+2}{3}
\end{align}
$
This solution is a complex root because it contains the imaginary unit $i$. This means the corresponding quadratic equation has no real solution, and therefore that the graph of this function has no $x$-intercepts.
## Factoring
When the square root method doesn't work because we can't easily isolate the square, we might try factoring next. It's worth noting that if we can factor, this method will always give us real solutions.
Going back to the notation $f(x) = ax^2 + bx + c$, factoring means finding two numbers $d$ and $e$ such that $d+e=b$ and $d \cdot e = a \cdot c$. This will allow us to factor and isolate the squared expression.
### $f(x)=x^2+7x+6$
Here, $a \cdot c = 6$ and $b=7$. We can pick the numbers $d=1$ and $e=6$, since $6+1=7$ and $6 \cdot 1 = 6$.
We can now start solving, rewriting the equation as follows:
$
\begin{align}
x^2+dx+ex+6=0
\\
x^2+1x+6x+6=0
\\
x(x+1)+6(x+1)=0
\\
(x+6)(x+1)=0
\end{align}
$
The zero product property tells us that if the result of a multiplication is zero, one of its terms must equal zero. Therefore there are two distinct solutions:
$
\begin{align}
x+6=0
\\
x=-6
\end{align}
$
and
$
\begin{align}
x+1=0
\\
x=-1
\end{align}
$
The $x$-intercepts of the function $f(x)=x^2+7x+6$ are $(-6,0)$ and $(-1,0)$.
### $g(x)=3x(x-4)-36$
W§e start by rewriting the equation as follows:
$
\begin{align}
3x(x-4)-36=0
\\
3x^2-12x-36=0
\\
3(x^2-4x-12)=0
\end{align}
$
We can now focus on solving what's inside the parentheses. From the zero product property, we know that one of the terms of the multiplication must equal $0$, and since $3 \ne 0$, we know our solutions don't depend on this factor.
Additionally, we pick the numbers $d = -6$ and $e = 2$, since $2 - 6 = -4$ and $-6 \times 2 = 1 \times -12$.
$
\begin{align}
x^2-4x-12=0
\\
x^2+dx+ex-12=0
\\
x^2-6x+2x-12=0
\\
x(x-6)+2(x-6)=0
\\
(x+2)(x-6)=0
\end{align}
$
Following, again, the zero product property, we find two distinct solutions:
$
\begin{align}
x+2=0
\\
x=-2
\end{align}
$
and
$
\begin{align}
x-6=0
\\
x=6
\end{align}
$
The $x$-intercepts of the function are therefore $(-2,0)$ and $(6,0)$.
### $h(x)=-2x^2-5x+12$
The negative sign in front of the quadratic means the graph will be a downward parabola.
We pick $d = 8$ and $e = -3$, since $8-3=5$ and $8 \times -3 = 12 \times -2$.
$
\begin{align}
-2x^2-5x+12=0
\\
-2x^2+8x-3x+12=0
\\
2x(4-x)+3(4-x)=0
\\
(2x+3)(4-x)=0
\end{align}
$
Following the zero product property gives us two distinct solutions:
$
\begin{align}
2x+3=0
\\
2x=-3
\\
x=-\frac{3}{2}
\end{align}
$
and
$
\begin{align}
4-x=0
\\
x=4
\end{align}
$
The zeros of our function are $x_1=-\frac{3}{2}$ and $x_2=4$.
### $i(x)=4x^2-12x+9$
We pick $d = -6$ and $e = -6$, since $-6-6=-12$ and $-6 \times -6 = 9 \times 4$.
$
\begin{align}
4x^2-12x+9=0
\\
4x^2-6x-6x+9=0
\\
2x(2x-3)-3(2x-3)=0
\\
(2x-3)^2=0
\\
\sqrt{(2x-3)^2}=\pm\sqrt{0}
\\
2x-3=\pm\sqrt{0}
\\
2x=3
\\
x=\frac{3}{2}
\end{align}
$
We see there is only one solution for this equation, and therefore one zero for this function, $\frac{3}{2}$, which means its graph touches the $x$-axis at exactly one point, $(\frac{3}{2}, 0)$.
## Completing The Square
Completing the square builds on the factoring method by making it possible when we cannot find two numbers $d$ and $e$ such that $d+e=b$ and $d \cdot e = a \cdot c$ with the equation as it is.
### $f(x)=x^2+4x-3$
### $g(x)=-2x^2+5x-7$
### $h(x)=x^2+\frac{2}{3}x-\frac{1}{3}$
## Quadratic Formula
The quadratic formula is derived from applying the [[Solving Quadratics#Completing The Square|completing the square]] method to a general equation in simple quadratic form. This yields the quadratic formula, which will allow us to express $x$ in terms of $a$, $b$ and $c$, and works all the time (however does not always yields a *real* solution).
### Proof
For $f(x)=ax^2+bx+c$,
$
\begin{align}
ax^2+bx+c=0
\\
a(x^2 + \frac{b}{a}x)+c=0
\end{align}
$
Following the completing the square method, we know we need to make factorable possible inside the parenthesis. To do so, we need to split $\frac{b}{a}x$ into two parts,$\frac{b}{2a}x$, add $\frac{b^2}{4a^2}$ inside the parentheses ($\frac{b}{2a}x \cdot \frac{b}{2a}x$), and remove $a\cdot\frac{b^2}{4a^2}$ outside the parenthesis to keep our equation balanced.
$
\begin{align}
a(x^2 + \frac{b}{2a}x + \frac{b}{2a}x + \frac{b^2}{4a^2})-a\cdot\frac{b^2}{4a^2}+c=0
\\
a(x(x+\frac{b}{2a}) + \frac{b}{2a}(x +\frac{b}{2a}))- a\cdot\frac{b^2}{4a^2} + c=0
\\
a(x +\frac{b}{2a})^2-\frac{a\cdot b^2}{4a^2}+c=0
\\
a(x +\frac{b}{2a})^2-\frac{b^2}{4a}+c=0
\\
a(x +\frac{b}{2a})^2=\frac{b^2}{4a} - c
\\
(x +\frac{b}{2a})^2=(\frac{b^2}{4a} - c)\cdot \frac{1}{a}
\\
(x +\frac{b}{2a})^2=\frac{b^2}{4a^2} - \frac{c}{a}
\\
(x +\frac{b}{2a})^2=\frac{b^2}{4a^2} - \frac{4ac}{4a^2}
\\
(x +\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}
\\
x +\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}
\\
x +\frac{b}{2a}=\frac{\pm\sqrt{b^2-4ac}}{\pm\sqrt{4a^2}}
\\
x +\frac{b}{2a}=\frac{\pm\sqrt{b^2-4ac}}{2a}
\\
x=-\frac{b}{2a} + \frac{\pm\sqrt{b^2-4ac}}{2a}
\\
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\end{align}
$
And voilà, we have the quadratic formula, which expresses $x$ in terms of $a$, $b$ and $c$.
$
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
$
### Application
When applying the quadratic formula:
- If the radicant is negative, you have two complex (and no real) solutions
- If the radicant is $0$, you have one real solution
- Otherwise you have two real solutions (and $x$-intercepts)
Considering $f(x)=3x^2-5x+1$,
$
\begin{align}
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\\
x=\frac{-(-5)\pm\sqrt{(-5)^2-(4)(3)(1)}}{2(3)}
\\
x=\frac{5\pm\sqrt{25-12}}{6}
\\
x=\frac{5\pm\sqrt{13}}{6}
\end{align}
$
Since the radicand is positive and not equal to zero, we know our equation has two real solutions, and that our function has two $x$-intercepts, $(0, \frac{5 - \sqrt{13}}{6})$ and $(0, \frac{5+\sqrt{13}}{6})$.
Note that it's best to wrap the numbers into parentheses as it's easy to make signs mistakes when plugging $a$, $b$ and $c$ into the formula.
### Fractions
Plugging in fractions to the quadratic formula will make your work harder.
Instead, you can get rid of fractions by multiplying both sides of the equation by the least common denominator.
Let's consider the following example:
$
\begin{align}
f(x)=\frac{1}{8}x^2-\frac{1}{4}x-2
\end{align}
$
#### Using Fractions
Trying to apply the quadratic formula right from the start gives us
$
\begin{align}
x=\frac{-(-\frac{1}{4}) \pm \sqrt{(-\frac{1}{4})^2- (4 \cdot \frac{1}{8} \cdot -2)}}{2 \cdot \frac{1}{8}}
\\
x=\frac{\frac{1}{4} \pm \sqrt{\frac{1}{16} - (-1)}}{\frac{1}{4}}
\\
x=(\frac{1}{4} \pm \sqrt{\frac{1}{16} + 1}) \times 4
\\
x=1 \pm 4 \sqrt{\frac{17}{16}}
\\
x=1 \pm 4 \frac{\sqrt{17}}{\sqrt{16}}
\\
x=1 \pm 4 \frac{\sqrt{17}}{4}
\\
x=1 \pm \sqrt{17}
\end{align}
$
#### Getting Rid of Fractions
However, if we start by simplifying our equation,
$
\begin{align}
\frac{1}{8}x^2-\frac{1}{4}x-2=0
\\
8 \cdot (\frac{1}{8}x^2-\frac{1}{4}x-2)=0 \cdot 8
\\
x^2-2x-16=0
\end{align}
$
We can now more easily apply the quadratic formula
$
\begin{align}
x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}
\\
x=\frac{-(-2) \pm \sqrt{(-2)^2 - (4 \times 1 \times -16)}}{2 \times 1}
\\
x=\frac{2 \pm \sqrt{4 - (-64)}}{2}
\\
x=\frac{2 \pm \sqrt{68}}{2}
\\
x=\frac{2 \pm \sqrt{4 \times 17}}{2}
\\
x=\frac{2 \pm (\sqrt{4} \cdot \sqrt{17})}{2}
\\
x=\frac{2 \pm 2 \sqrt{17}}{2}
\\
x=\frac{2 \cdot (1 \pm \sqrt{17})}{2}
\\
x=1 \pm \sqrt{17}
\end{align}
$
Both approaches yield the same result. However, plugging fractions into the quadratic formula is more error-prone than the counterpart, which should generally be preferred.
## Example Application
Imagine you're playing football with your friends, and during the game the ball got stuck in a tree. You're throwing rocks at the ball in the hope of getting it down. The ball is stuck 7 meters high in the tree, and the trajectory of the rocks you're throwing at it to try and get it unstuck is a downward parabola and can be described by the function
$
\begin{align}
f(x) = -9x(x+2)-3
\end{align}
$
with you standing at $y=0$. Will you be able to touch the ball?
To solve this, instead of solving for $f(x)=0$, we can solve for $f(x)=7$.
We start by writing our polynomial in order to make our work easier.
$
\begin{align}
f(x)=-9x^2-18x-3
\end{align}
$
We can't isolate the power of two just yet, and can't use the square root method as is.
We can't resort to factoring either, as there are no numbers $d$ and $e$ such that $d+e=b$ and $d \cdot e = a \cdot c$.
Now that we understand the quadratic formula, we'd like to leverage it. However it will give us the *zero* of the function, when instead we want the *seven* of the function, so to speak. We want to know if our rock will reach a height of seven meters, which translates with our function crossing or touching $f(x)=7$. This mean the quadratic equation we actually want to solve is
$
\begin{align}
-9x^2-18x-3=7
\end{align}
$
Let's take an intermediary step and tweak the equation so that applying the quadratic formula will give us our answer. Rewriting it like so applies a [[Graphing Functions#Vertical Shift|vertical shift]]. It's like we're moving the $x$-axis higher to fit our question.
$
\begin{align}
-9x^2-18x-3-7=0
\\
-9x^2-18x-10=0
\end{align}
$
Now that we're solving for zero, let's apply the quadratic formula.
$
\begin{align}
x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}
\\
x=\frac{-(-18) \pm \sqrt{(-18)^2- (4 \times -9 \times -10)}}{2 \times -9}
\\
x=\frac{18 \pm \sqrt{324- 360}}{-18}
\\
x=\frac{18 \pm \sqrt{-36}}{-18}
\\
x=\frac{18 \pm i\sqrt{36}}{-18}
\end{align}
$
At this point we see the radicand is a negative number, meaning the quadratic has no real solutions, and therefore no $x$-intercept for our transformed equation.
Unfortunately, you won't be able to knock the ball down by throwing rocks at it.