Solving Quadratic Functions

Published: Aug 20th, 2020

Learning Objectives
By the end of this section, you should be able to:
1. Solve a function by factoring
2. Solve a function using the quadratic equation
3. Solve a quadratic inequality

In this lesson, we will discuss something you may have covered extensively in the tenth grade: quadratic functions. These functions are in the form \(f(x) = ax^2 + bx + c\), where \(a\)\(b\), and \(c\) are all real numbers. 

When you are asked to solve an equation, quadratic or not, this typically means to find values of \(x\) for which the function equals some other value. In this case, we are looking for values of \(x\) for which the function \(f(x)\) is equal to zero. These points are called the zeroes, or roots of a function. The graphic below shows how we can find the root for an unknown function by looking at its graph:

quadratic function with roots (-5,0) and (5,0)
Looking at its graph, this quadratic function has roots at (-5,0) and (5,0)

The graph of this function takes on a shape called a parabola. The shape of this parabola can either be described as concave upwards (yielding the smiley face smiley shape) or downwards (frowny face sad). Believe it or not, we encounter the parabolic motion (motion in a parabolic shape) very often. This is because all objects on the earth's surface experience a constant force of gravity, \(F_g\) which is related to the object's mass by a constant called the acceleration due to gravity \(g\), with magnitude 9.81 \(m/s^2\). One example of parabolic motion is shown in the graphic below:

parabolic motion example
We can model the vertical position of the calculator with respect to the ground using the equation \(h(x) = 12 - 9.8x^2\), where  \(x\) is the time in seconds.

Introductions and silliness aside, let's go over three ways we can solve a quadratic equation:

Solving by Factoring

When a function can be factored, we can find its roots by factoring then solving each root for \(x\).

Factor \(x^2 + 2x + 1\)

Hopefully this one's pretty easy. We need two numbers that multiply to 1 and add to 2: 1 and 1! Writing this in factored form, we get: 

\(x^2+2x+1 = (x+1)^2\)

Recall that solving means to find the zeroes, or roots of the equation. In this case, we set \((x+1)^2 = 0\), which gives \(x+1 = 0\), or \(x=-1\).

And now an \(a>1\) example:

Factor \(2x^2+x-3\)

In this case, we are looking for an expression in the form \((ax+b)(cx+d)\) where \(ac = 2\)\(bd = -3\), and \(ad+bc = 1\).  By trial and error, we get \((2x+3)(x-1)\).

To solve, we set \((2x+3) = 0 \) and \((x-1) = 0\), giving \(x = \frac{-3}{2}\) and \(x=1\).

Solving with the Quadratic Equation

Now, what if the equation can't be factored? Well, luckily there's the quadratic equation:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\), where \(a\)\(b\), and \(c\) are from the quadratic form of the equation: \(ax^2+bx+c\)

Let's try a few examples:

Factor \(x^2 – 3x + 1 = 0\)

Well, let's at least try factoring without the quadratic equation... What two numbers multiply to 1 and add to -3? 

Hmmmm..... hmmmmmmmmm...... well, if you said "there aren't any," you'd be right... sort of!

Let me explain. Let's try plugging this equation into the quadratic formula, where \(a=1\)\(b=-3\), and \(c=1\):

\(x = {-(-3) \pm \sqrt{(-3)^2-4(1)(1)} \over 2(1)}\)

\(x = {3 \pm \sqrt{9-4} \over 2}\)

\(x = {3 \pm \sqrt{5} \over 2}\)

So, there are two real number solutions: \(x = {3 + \sqrt{5} \over 2}\)and \(x = {3 - \sqrt{5} \over 2}\). Bet you wouldn't have thought of those on your own, eh?

Anyway, let's try another example:

Common factor the 5: \(5(x^2-2x+2)\)

Two numbers that multiply to 2 and add to -2... can't think of any... quadratic it is!

\(x = {-(-2) \pm \sqrt{(-2)^2-4(1)(2)} \over 2(1)}\)

\(x = {2 \pm \sqrt{4-8} \over 2}\)

\(x = {2 \pm \sqrt{-4} \over 2}\)

W-w-what?! The square root of a negative number! Preposterous! Well, not quite. 

If you remember from Lesson #4: Types of Numbers we have a name for these types of numbers: Complex Numbers

This means that we cannot plot these numbers in the real number plane, represented by \(\mathbb{R}\). Instead, these are part of a larger group of numbers that includes the real numbers... the complex numbers \(\mathbb{C}\). Fancy letters indeed, but all you need to know is that \(i = \sqrt{-1}\). Given this, let's factor out \(i\) from the equation:

\(x = {2 \pm i\sqrt{4} \over 2}\)

\(x = {2 \pm 2i \over 2}\)

\(x = 1 \pm i\), or \(1+i\) and \(1+-i\). Note that \(-i = -\sqrt{-1} \neq 1\)

That's as simple as it gets!

Now, like the cartoon illustrated, you might be wondering "When am I ever going to use this in my life, anyway?" Well, that's a good segue for the next part of this lesson!

Solving Quadratic Inequalities

The last time we looked at quadratic inequalities (Lesson #10: Inequalities) I sort of condescendingly addressed the subject. 

Well, the sass only came because we weren't quite equipped to answer those questions quite yet. Now, however, we are!

Let's have a crack at it, shall we?

When is \(4x^2-16x+4 \gt 0\)

Common factoring the 4, we get: \(x^2 - 4x + 1\). When is this function greater than zero?

Let's solve. We need two numbers that multiply to 1 and add to -4. Looks like a job for the quadratic equation.

Letting \(a=1\)\(b=-4\), and \(c=1\):

\(x = {-(-4) \pm \sqrt{(16)-4(1)(1)} \over 2(1)}\)

\(x = {4 \pm \sqrt{12} \over 2}\)=\(\frac{4\pm 2 \sqrt{3}}{2}\)=\(2\pm\sqrt{3}\)

Now we have two important things: our zeroes, and an idea of the function's shape. How do we know the latter?

Well, looking at the equation, we see that \(a>0\). This is a smiley face function. Because we have its zeroes, we know that anything between these zeroes must be negative, because the function hasn't crossed \(y=0\) at that point yet! You can prove it for yourself by plugging in any number between \(2 - \sqrt{3}\)and \(2 + \sqrt{3}\) into \(x^2 - 4x + 1\). The result should be negative. If not, please call (678) 999-8212. 

Now, if everything between those numbers is less than zero, and those numbers are the zeroes of the function (the x values where \(y=0\)), then everything on either side of the zeroes must be positive! In math terms:

\(x \lt 2 - \sqrt{3}\) or \(x \gt 2 + \sqrt{3}\).

Applications of Quadratic Functions

Let's say you're a lemonade salesperson.

lemons
A classy gent.
You sell lemonade for $1.50 a cup. On average, you sell 100 cups each day. You're trying to find out the best possible price, so you've been doing some market research. You notice that, for every $0.10 you increase the price, you lose 2 customers. What is the best price at which you can maximize revenue?

Let's write this in equation form:

\((1.50 + 0.10x)(100-2x)\).

The function represents our revenue. For every $0.10 added to the price, represented by \(x\), we lose 2 customers. 

This type of problem is called an optimization problem. We're trying to find the best possible price at which we can maximize the revenue. 

Now, let's plot this equation (you won't have to do this every time, but it's helpful to understand the principle behind these problems):

Looking at this curve, we see that \(a<0\); it's a frowny-face parabola. There are two roots: one negative and one positive.

Recall that this function represents our revenue. As businesspeople, we're trying to maximize our revenue. In other words, we want to find the value of \(x\) for which the parabola is at its maximum; the extreme value of this function.

Now, how can we do that? Well, there are two ways. First, we can solve the function (which is luckily already factored!):

\(1.50 + 0.10x = 0\). Multiply both sides by 10

\(15 + x = 0 \)

\(x = -15\)

\(100-2x=0\)

\(x=50\).

These x values represent each root of the parabola. As you may recall from Lesson #7: Four Basic Functions, all functions are symmetric about a vertical axis. Given that we have two points on either edge of this symmetrical figure that share the same \(y\) value (zero), then we can take the average of these two points to get the midpoint:

\(\frac{50-15}{2}\)\(\frac{35}{2}\), or 17.5

In other words, to maximize our revenue, we can increase our price by 17.5(0.1) = $1.75, to $3.25 (1.75+1.50), and still retain 100-2(17.5) = 65 customers. 

Now, you may be wondering about the other way. Well, I don't really like it myself because it involves memorizing, and that's not what math's about!

But, if you want, the formula to find the midpoint is \(\frac{-b}{2a}\). Expanding our function, we get:

\(150 + 7x - 0.2x^2\)

Where \(a = -0.2\)\(b=7\), and \(c=150\).

\(\frac{-7}{2(-0.2)}\)\(\frac{-70}{-4}\)\(17.5\). Same result!

See? There are some applications to this stuff. With that, let's move on to some practice problems, shall we?

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