A quadratic is a second-degree [[Polynomials|polynomial]]. In other words, a quadratic is a [[Functions|function]], 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 [[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.
It merely consists of an additional step to rewrite the terms in a way that makes factoring possible, introducing the missing numbers $d$ and $e$ such that $d+e=b$ and $d \cdot e = a \cdot c$.
We start by factoring out $a$ if $a\ne1$, we split $b$ in two equal parts, $\frac{b}{2}x+\frac{b}{2}x$, add $\frac{b}{2}^2$ in the factored part and remove $a \cdot \frac{b}{2}^2$ afterwards.
>To prove a quadratic function where we reverse signs has the same exact $x$-intercepts, that $f(x)$ is now $-f(x)$, and more generally that the new function is but a reflection of $f(x)$ across the $x$-axis, let us consider the following equality:
$
\begin{align}
ax^2+bx+c=-ax^2-bx-c
\\
2ax^2+2bx^2+2c=0
\\
2(ax^2+bx+c)=2 \cdot 0
\\
ax^2+bx^2+c=0
\end{align}
$
>Since $ax^2+bx+c=-ax^2-bx-c$ is equivalent to $ax^2+bx^2+c=0$, we see that $ax^2+bx^2+c=0$ has the exact same solutions as $-ax^2-bx-c=0$.
Therefore, regardless of whether we reverse the signs within a function or not, we will be able to find the correct $x$-intercepts.
The following examples will make this strategy much clearer.
### $f(x)=x^2+4x-3$
Here, $a=1$ so we need not factoring it out.
$
\begin{align}
x^2+4x-3=0
\\
x^2+2x+2x-3=0
\\
(x^2+2x+2x+4)-4-3=0
\end{align}
$
Notice how we've rewritten the equation and isolated a term on which we can easily apply the factoring method (between parentheses). We've kept our equation balanced, because we've added and removed the same number from the left side of our equation.
$
\begin{align}
(x(x+2)+2(x+2))-7=0
\\
(x+2)^2-7=0
\\
(x+2)^2=7
\\
x+2=\pm\sqrt{7}
\\
x=-2\pm\sqrt{7}
\end{align}
$
There are two real solutions since the radicand is positive, $x=-2+\sqrt{7}$ and $x=-2-\sqrt{7}$.
### $g(x)=-2x^2+5x-7$
Here we can't apply factoring directly. We want $a=1$, not a negative number.
$
\begin{align}
-2x^2+5x-7=0
\\
-1\cdot(2x^2+5x-7)=0\cdot-1
\\
2x^2-5x+7=0
\\
2(x^2-\frac{5}{2}x)+7=0
\end{align}
$
Here, $b=-\frac{5}{2}$, so we need to add $b=-\frac{5}{2}^2$ within the parentheses to get a factorable expression, and remove $b=-2\cdot\frac{5}{2}^2$ outside to keep our equation in balance.
$
\begin{align}
2(x^2-\frac{5}{4}x-\frac{5}{4}x+(-\frac{5}{4}^2))-2(-\frac{5}{4}^2)+7=0
\\
2(x^2-\frac{5}{4}x-\frac{5}{4}x+\frac{25}{16})-2(\frac{25}{16})+7=0
\\
2(x(x-\frac{5}{4})-\frac{5}{4}(x-\frac{5}{4}))-\frac{25}{8}+\frac{56}{8}=0
\\
2(x-\frac{5}{4})^2+\frac{31}{8}=0
\\
2(x-\frac{5}{4})^2=-\frac{31}{8}
\\
(x-\frac{5}{4})^2=-\frac{31}{8}\cdot\frac{1}{2}
\\
(x-\frac{5}{4})^2=-\frac{31}{16}
\\
x-\frac{5}{4}=\pm\sqrt{-\frac{31}{16}}
\\
x=\frac{5}{4}\pm i\sqrt{\frac{31}{16}}
\end{align}
$
Again, the square root of a negative number yields an imaginary number, which means this equation has two complex solutions and the graph of the function has no $x$-intercepts.
### $h(x)=x^2+\frac{2}{3}x-\frac{1}{3}$
$
\begin{align}
x^2+\frac{2}{3}x-\frac{1}{3}=0
\\
(x^2+\frac{1}{3}x+\frac{1}{3}x)-\frac{1}{3}=0
\\
(x^2+\frac{1}{3}x+\frac{1}{3}x+\frac{1}{9})-\frac{1}{9}-\frac{3}{9}=0
\\
(x(x+\frac{1}{3})+\frac{1}{3}(x+\frac{1}{3}))-\frac{4}{9}=0
\\
(x+\frac{1}{3})^2-\frac{4}{9}=0
\\
(x+\frac{1}{3})^2=\frac{4}{9}
\\
x+\frac{1}{3}=\pm\sqrt{\frac{4}{9}}
\\
x=-\frac{1}{3}\pm\sqrt{\frac{4}{9}}
\end{align}
$
The two $x$-intercepts of $h(x)$ are located at $(-\frac{1}{3}-\sqrt{\frac{4}{9}},0)$ and $(-\frac{1}{3}+\sqrt{\frac{4}{9}},0)$.
## Quadratic Formula
The quadratic formula is derived from applying the [[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, $(\frac{5 - \sqrt{13}}{6},0)$ and $(\frac{5+\sqrt{13}}{6},0)$.
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 [[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.
## Intersection of Functions
We should now be able to see that what we've been doing so far is finding the intersection of functions. Indeed, what is an $x$-intercept, but the intersection of a function $f(x)$ and another defining the $x$-axis itself, $g(x)=0$?
This is exactly what we've done in our previous example, except we were thinking of it in terms of 'moving up' the $x$-axis instead of finding the intersection of $f(x) = -9x(x+2)-3$ and $g(x)=7$.
Generalizing that idea, we can find the intersection of two functions by setting one equal to the other (instead of setting the equation equal to $0$ like we've been doing so far).
### $f(x)=x^2+6x+3$ and $g(x)=3$
Let us consider $f(x)=x^2+6x+3$ and $g(x)=3$. We start by getting everything on one side in order with a first term that's positive.
$
\begin{align}
x^2+6x+3=3
\\
x^2+6x=0
\end{align}
$
Conveniently, we're now let with a quadratic equation to solve, which we already know well how to solve.
$
\begin{align}
x(x+6)=0
\end{align}
$
Following the zero product property, we find two solutions $x=-6$ and where $x=0$. This means our functions intersect when we plug either of those numbers into them.
Thus, the points of intersection themselves are defined in terms of $(x,f(x))$ or $(x,g(x))$. Here, they are $(-6, 3)$ and $(0, 3)$.
### $f(x)=-2x^2+1$ and $g(x)=3x+2$
$
\begin{align}
-2x^2+1=3x+2
\\
-2x^2+1-3x-2=0
\\
-2x^2-3x-1=0
\\
2x^2+3x+1=0
\\
2x^2+2x+1x+1=0
\\
2x(x+1)+1(x+1)=0
\\
(2x+1)(x+1)=0
\end{align}
$
By the zero product property, we find two real solutions.
$
\begin{align}
2x+1=0
\\
x=-\frac{1}{2}
\end{align}
$
and
$
\begin{align}
x+1=0
\\
x=-1
\end{align}
$
This means our functions intersect for $x=-1$ and $x=-\frac{1}{2}$.
The points of intersection are therefore $(-1, f(-1))=(-1,-1)$ and $(-\frac{1}{2},f(-\frac{1}{2}))=(-\frac{1}{2},0.5)$.
>⚠️ If you have a calculator, say the TI-84, like I do, you need to wrap numbers in parentheses before squaring them. Otherwise the negative sign gets applied *after* the number have been squared, and $-1^2$=-1. Save yourself the headache.
### $f(x)=x^2-x+1$ and $g(x)=2x^2-3x-14$
$
\begin{align}
x^2-x+1=2x^2-3x-14
\\
0=2x^2-3x-14-x^2+x-1
\\
x^2-2x-15=0
\\
x^2+3x-5x-15=0
\\
x(x+3)-5(x+3)=0
\\
(x+3)(x-5)=0
\end{align}
$
By the zero product property, we're able to find the two solutions are $x=-3$ and $x=5$.
Our two points of intersection are $(-3,13)$ and $(5,21)$.
## Substitutions
Solving quadratics doesn't apply only to polynomials composed of variables that resolve to plain values; you can apply the same techniques to quadratic polynomials composed of expressions.
To do so, we add an intermediary step and declare a variable that makes the calculation easier.
### $g(x)=(3x+4)^2-6(3x+4)+9$
$
\begin{align}
(3x+4)^2-6(3x+4)+9=0
\end{align}
$
Let us declare a variable $u=3x+4$.
$
\begin{align}
u^2-6u+9=0
\\
u^2-3u-3u+9=0
\\
u(u-3)-3(u-3)=0
\\
(u-3)^2=0
\\
u-3=\pm\sqrt{0}
\\
u=3
\end{align}
$
We can now substitute $u$ with its actual value, $3x+4$.
$
\begin{align}
3x+4=3
\\
3x=-1
\\
x=-\frac{1}{3}
\end{align}
$
Our equation has one real solution. The graph of this function touches the $x$-axis at $(-\frac{1}{3},0)$.
### $f(x)=3x^4-2x^2-1$
From the power of $4$ we get the intuition we should get $4$ solutions.
$
\begin{align}
3x^4-2x^2-1=0
\end{align}
$
Let us declare the variable $u=x^2$ and rewrite our polynomial.
$
\begin{align}
3u^2-2u-1=0
\\
3(u^2-\frac{1}{3}u-\frac{1}{3}u+\frac{1}{9})-\frac{1}{3}-1=0
\\
3(u(u-\frac{1}{3})-\frac{1}{3}(u-1))-\frac{4}{3}=0
\\
3(u-\frac{1}{3})^2=\frac{4}{3}
\\
(u-\frac{1}{3})^2=\frac{4}{9}
\\
u-\frac{1}{3}=\pm\sqrt{\frac{4}{9}}
\\
u=\frac{1}{3}\pm\frac{\sqrt{4}}{\sqrt{9}}
\\
u=\frac{1}{3}\pm\frac{2}{3}
\end{align}
$
Now we've isolated our variable $u$, we can expand it.
$
\begin{align}
x^2=\frac{1}{3}\pm\frac{2}{3}
\\
x=\pm\sqrt{\frac{1}{3}\pm\frac{2}{3}}
\end{align}
$
There are two real solutions, $-1$ and $1$ and two complex ones, $-i{\frac{\sqrt{3}}{3}}$ and $i{\frac{\sqrt{3}}{3}}$.
This means we have two $x$-intercepts, located at $(-1,0)$ and $(1,0)$.
### $h(x)=x^\frac{2}{3}-7x^\frac{1}{3}+10$
Using the power of a power rule ($(x^a)^b=x^{a \cdot b}$), let $u=x^{\frac{1}{3}}$.
$
\begin{align}
u^2-7u+10=0
\\
u(u-5)-2(u-5)=0
\\
(u-2)(u-5)=0
\end{align}
$
Using the zero product property, we solve the first expression.
$
\begin{align}
u-2=0
\\
x^{\frac{1}{3}}-2=0
\\
x^{\frac{1}{3}}=2
\\
(x^{\frac{1}{3}})^3=2^3
\\
x=8
\end{align}
$
We now solve the second expression.
$
\begin{align}
u-5=0
\\
x^{\frac{1}{3}}-5=0
\\
x^{\frac{1}{3}}=5
\\
(x^{\frac{1}{3}})^3=5^3
\\
x=125
\end{align}
$
We get two real solutions and $x$-intercepts.
>Fractional exponents are but a way to represent powers and roots together:
$
\begin{align}
\sqrt[3]{x}=x^{\frac{1}{3}}
\\
(\sqrt[3]{x})^3=(x^{\frac{1}{3}})^3=x
\end{align}
$
### $f(x)=x+\sqrt{x}-20$
Building on the previous example, let $u=\sqrt{x}=x^{\frac{1}{2}}$.
$
\begin{align}
u^2+u-20=0
\\
u^2-4u+5u-20=0
\\
u(u-4)+5(u-4)=0
\\
(u+5)(u-4)=0
\end{align}
$
Using the zero product property, we solve the **second** expression.
$
\begin{align}
u-4=0
\\
\sqrt{x}=4
\\
x=16
\end{align}
$
However, we can't solve the first expression, because it's not possible for $\sqrt{x}$ to yield $-5$.
$x=-5$ is outside the domain of our function.
Because it has a square root in it, not only is $-5$ but also all negative inputs, as $\sqrt{x}$ is only defined for $x\geq 0$.
The domain of the function is $[0,\infty)$.
We find only one real solution and $x$-intercept.
### $h(x)=-2y^{-2}-11y^{-1}+40$
Remembering a negative exponent is the reciprocal of a number,
$
\begin{align}
x^{-a}=\frac{1}{x^a}
\end{align}
$
let $u=y^{-1}$.
$
\begin{align}
-2u^2-11u+40=0
\\
2u^2+11u-40=0
\\
2u^2-5u+16u-40=0
\\
u(2u-5)+8(2u-5)=0
\\
(u+8)(2u-5)=0
\end{align}
$
Let's solve the first expression.
$
\begin{align}
u+8=0
\\
u=-8
\\
y^{-1}=-8
\\
(y^{-1})^{-1}=-8^{-1}
\\
y=-\frac{1}{8}
\end{align}
$
Solving the second expression,
$
\begin{align}
2u-5=0
\\
u=\frac{5}{2}
\\
(y^{-1})^{-1}=(\frac{5}{2})^{-1}
\\
y=\frac{2}{5}
\end{align}
$
We get two real solutions and $x$-intercepts.
## Graphing
Quadratics functions can be wide or narrow, upward or downward parabolas and be subjected to transformations, but they will always be parabolas with a vertical axis of symmetry.
Understanding what the graph of a quadratic function looks like allows us to build an understanding of the relationship between its input and output, and can sometimes save us some steps.
For instance, if you know the lowest point of the parabola is above the $x$-axis and that it is an upward-opening parabola, you know you don't need to look for $x$-intercepts, because they won't be any real solutions.
To build a visual representation of a quadratic function, we follow these steps:
1. Define the parabola orientation.
This is exclusively dependent on the sign of the quadratic term. If it's positive, the parabola will be upward opening. If it's negative, it's a downward opening parabola.
2. Find the vertex of the parabola.
This is the minima or the maxima, i.e. the highest or lowest point on the parabola through which passes the axis of symmetry. To find this point, we use the *vertex formula*, which is derived from the *quadratic formula*: $(-\frac{b}{2a}, f(-\frac{b}{2a}))$
3. Find the $y$-intercept.
There will always be one. To do so, we input $x=0$, which gives us $c$. Indeed, $a\cdot0^2+b\cdot0+c=c$. So the $y$-intercept can be found at $(0,c)$. We can then reflect this point across the vertical axis of symmetry (the line parallel to the $y$-axis that goes through the vertex of the parabola). We now have three points defining our parabola, which gives us a rough understanding of its shape.
4. If necessary, we can now figure out what the $x$-intercepts are.
Understanding what the graph of the function looks like will also allow us to do a sanity check on out solutions.
Let's now apply this methodology to a couple of functions.
### $f(x)=2x^2+8x+5$
1. This is an upward opening parabola.
2. $-\frac{b}{2a}=-\frac{8}{2\times2}=-2$ and $f(-2)=-3$. The vertex is at $(-2,-3)$.
3. $f(0)=5$. Our $y$-intercept is at $(0,5)$, and its corresponding point across the axis of symmetry will be at $(-4,5)$.
4. $x=-2\pm\frac{\sqrt{6}}{2}$. We find two real solutions and $x$-intercepts.
### $g(x)=-2x^2-4x+2$
1. This is a downward opening parabola.
2. $-\frac{b}{2a}=-\frac{-4}{2\times-2}=-1$ and $g(-1)=4$. The vertex is at $(-1,4)$.
3. $g(0)=2$. Our $y$-intercept is at $(0,2)$, and its corresponding point across the axis of symmetry will be at $(-2,2)$.
4. $x=-1\pm\sqrt{2}$. We find two real solutions and $x$-intercepts.
### $f(x)=-4x^2-4x-1$
1. This is a downward opening parabola.
2. $-\frac{b}{2a}=-\frac{-4}{2\times -4}=-\frac{1}{2}$ and $f(-\frac{1}{2})=0$. The vertex is at $(-\frac{1}{2},0)$.
3. $f(0)=-1$. Our $y$-intercept is at $(0,-1)$, and its corresponding point across the axis of symmetry will be at $(-1,-1)$.
4. The vertex already gave us the zero of the function. We therefore know there are two identical real solutions, meaning the parabola touches the $x$-axis when $x=-\frac{1}{2}$.
### $g(x)=3x^2+2x+5$
1. This is an upward opening parabola.
2. $-\frac{b}{2a}=-\frac{2}{2\times3}=-\frac{1}{3}$ and $g(-\frac{1}{3})=\frac{14}{3}$. The vertex is at $(-\frac{1}{3},\frac{14}{3})$.
3. $g(0)=5$. Our $y$-intercept is at $(0,5)$, and its corresponding point across the axis of symmetry will be at $(-\frac{2}{3},5)$.
4. Since this is an upward opening parabola and the vertex is located above the $x$-axis, there are no real solutions to be found.
### Challenge
For a parabola with a vertex of $(1,-5)$ and a $y$-intercept of $-3$, find the function such that $f(x)=a(x+h)^2+k$.
Knowing the coordinates of the vertex gives us information about both the horizontal and vertical shifts applied to the quadratic function, since we know those will always be symmetric, that their axis of symmetry will always pass through the vertex and that it will always be parallel to the $y$-axis.
The $x$-coordinate of $1$ corresponds to a *right shift*, while the $y$-coordinate of $-5$ corresponds to *down shift*. Building on our knowledge of [[Functions#Transformations|transformations]], we see that $h=-1$ and $k=-5$.
$
\begin{align}
f(x)=a(x-1)^2-5
\end{align}
$
How can we now go about finding $a$? Using the knowledge we have that the $y$-intercept is located at $(0,-3)$, we plug in $x=0$ and solve the following:
$
\begin{align}
f(x)=a(x-1)^2-5
\\
-3=a(0-1)^2-5
\\
2=a\times1
\\
a=2
\end{align}
$
Therefore,
$
\begin{align}
f(x)=2(x-1)^2-5
\end{align}
$
## Inequalities
Quadratic inequalities combine the idea of a quadratic function with the idea of something being inferior or superior to something else. They produce *truth statements*.
This is the same type of function inequalities used in engineering applications to determine if an electrical signal should be interpreted as a $0$ or a $1$. This is how *booleans* are built. Such applications, however, are much simpler in that they're based on linear inequalities.
>⚠️ Don't forget that while solving inequalities, dividing by a negative number requires us flipping the inequality. So if you reverse the signs to be able to solve an inequality, you should reverse the inequality as well. For instance, $-ax^2-bx-c\leq0$ can be rewritten as $ax^2+bx+c\geq0$, because you multiply both sides of the inequality by $-1$.
### $x^2-3x-10 \leq 0$
This inequality can be interpreted as finding the values of $x$ such that $f(x)\leq0$, that is the values of $x$ for which $f(x)$ is *below* the $x$-axis.
$
\begin{align}
x^2-3x-10 \leq 0
\\
x^2-5x+2x-10 \leq 0
\\
x(x-5)+2(x-5) \leq 0
\\
(x+2)(x-5) \leq 0
\end{align}
$
Following the zero product property, we find two solutions, $x \leq -2$ and $x \leq 5$.
This means that $f(x) \leq 0$ within the interval $[-2,5]$.
It's worth noting that because of the nature of parabolas, quadratic inequalities always produce alternative truth statements. Jumping ahead, we see how this is also the case for trigonometric and other cyclic functions.
Here, our alternative truth statements are $false$, $true$, $false$.
### $2x^2 \leq 5x + 3$
Let's take a look back at our previous example first. If we see the $x$-axis as a function such that $g(x)=0$, then we were looking for values of $x$ such that $f(x)$ was below that function.
This example, $2x^2 \leq 5x + 3$, builds on this idea and compares two functions.
Let us define $f(x)=2x^2$ and $g(x)=5x+3$.
What we're solving for is the range of values of $x$ for the functions on both sides of the inequality such that $f(x) \leq g(x)$. In other words, we're asking where the parabola defined by the quadratic function $f(x)$ is below the line defined by the linear function $g(x)$.
$
\begin{align}
2x^2 \leq 5x + 3
\\
2x^2 - 5x - 3 \leq 0
\\
2x^2-6x+1x-3 \leq 0
\\
2x(x-3)+1(x-3) \leq 0
\\
(2x+1)(x-3) \leq 0
\end{align}
$
Following the zero product property, we find two real solutions, $x \leq -\frac{1}{2}$ and $x \leq 3$.
This means that for the values of $x$ contained within the interval $[-\frac{1}{2},3]$, $f(x)\leq g(x)$.
This truth statement does not hold outside this interval and will be resolved as $false$.
### $x(x+1)\gt 20$
$
\begin{align}
x(x+1)\gt 20
\\
x^2+x\gt 20
\\
(x^2+\frac{1}{2}x+\frac{1}{2}x+\frac{1}{4})-\frac{1}{4} \gt 20
\\
(x+\frac{1}{2})^2 \gt 20 + \frac{1}{4}
\\
x \gt -\frac{1}{2} \pm \sqrt{\frac{81}{4}}
\\
x \gt \frac{-1 \pm 9}{2}
\end{align}
$
We find two real solutions, $x \gt -5$ and $x \gt 4$.
Thus, the statement holds true for values of $x$ within the interval $[-5,4]$.
### $4x^2+9\lt 6x$
$
\begin{align}
4x^2+9 \lt 6x
\\
4x^2-6x+9 \lt 0
\\
4(x^2-\frac{6}{8}x-\frac{6}{8}x+\frac{36}{64})-\frac{36}{16}+\frac{144}{16} \lt 0
\\
(x-\frac{6}{8})^2 \lt -\frac{108}{16} \cdot \frac{1}{4}
\\
x-\frac{6}{8} \lt \pm \sqrt{-\frac{108}{64}}
\\
x \lt \frac{6}{8} \pm \frac{\sqrt{-108}}{8}
\\
x \lt \frac{6 \pm i\sqrt{108}}{8}
\end{align}
$
This statement will always be $false$, as there are no real solutions.
The functions on the left of the inequality never crosses the function on its right.
## Example Application
Imagine you have $3000$ meters of fence, and you want to find out what is the maximum area $A$ that you can enclose within that fence.
Since we want to maximize the area but are given a limit in terms of the perimeter, let's solve the perimeter constraint first for the variable we want to maximize. Since we'll be working with a rectangle, we label one side length $x$ and the other $y$.
The formula for the perimeter of a rectangle is $P=2x+2y$. The maximum amount of perimeter we have is $3000$.
Therefore, choosing arbitrarily to solve for $y$,
$
\begin{align}
2x+2y=3000
\\
2y=3000-2x
\\
y=1500-x
\end{align}
$
Now that we expressed one of our variables in terms of the other and incorporated the limit in this function, we can rewrite the area formula $A=x \cdot y$.
$
\begin{align}
A=x \cdot (1500-x)
\\
A=-x^2+1500x
\end{align}
$
We now have a function representing how the area we can surround with our $3000$ meters of fence grows alongside the length $x$ of two sides.
This is a quadratic function. To find the optimal value of $x$ that will allow us to get the biggest possible area, we look for the vertex.
$
\begin{align}
\frac{-b}{2a}=\frac{-1500}{2\cdot -1}=750
\end{align}
$
By picking a side length for $x$ of $750$ meters, we ensure we'll get the biggest possible area.
Now coming back to the formula that expresses $y$ in terms of $x$, we see that the $y$ value must be
$
\begin{align}
1500-x=1500-750=750
\end{align}
$
By picking a land that's $750 \times 750$ meters, we ensure we'll use up all of the fence we have available, thus reaching the maximum area we could, which is $A=750 \times 750=562500$ square meters.