Quadratic Equations

So far we have only looked at linear equations. What this means in this context is none of the variables have had an exponent (other than, technically, 1). We will now start to look at equations that have a power of 2 in them.

Let's start by graphing a basic quadratic equation. What we will consider is the most basic equation possible \(y = x^2\)

The important thing to note about this graph is that it is symmetrical. Consider drawing a horizontal line anywhere above the x-axis. Other than at 0, we can easily see that this line will pass through two points. This is really significant - this means that for any solution that is not 0, this simple equation has two solutions. Consider \(x^2 = 4\). Obviously, the solution is \(x = \sqrt{4}\). But what is \(\sqrt{4}\)? There are two possible values: both 2 and -2. Generally, quadratic equations have two solutions, a positive one and a negative one

Let's look at the more general quadratic equation

\[y = ax^2 + bx + c\]

\(a\) is the scaling factor. The larger \(a\) the steeper the rise of the curve. If \(a\) is negative then the graph will be mirrored in the x-axis. Below is a plot of \(y = -x^2\).

\(b\) moves the graph from left to right and down. If \(b\) is positive then the graph will overall shift to the left, and the lowest point will shift down if \(a\) is positive. If \(a\) is negative a positive \(b\) will shift the graph to the left and up. These statements are all reversed for a negative \(a\). The next graph shows \(y = x^2 + 2x\) and \(y = -x^2 - 4x\).

Finally \(c\) is totally analogous to the c we met before in the linear section - it simply shifts the graph up and down. The next graph shows \(y = x^2 + 5\) and \(y = -x^2 - 3x - 2\).

Now that we have an understanding of how a quadratic equation behaves let's explore methods of solving it. First, let's work on expanding and factorising brackets. Consider the following

\[y = (x + a)(x + b)\]

Expanded this becomes

\[y = x^2 + ax + bx + ab\]

Notice how after expanding this bracket we have a quadratic equation. Expanding the brackets is very easy and mechanical, and in general

\[(a+b)(c+d) = ac + bc + ad + bd\]

Some people remember this with the acronym FOIL. You need to multiply the First, then the Inside, then the Outside, then the Last. However, if the brackets have more than two terms each then this breaks down, so I try to think about multiplying every term in the left bracket with every term in the right bracket.

Expand the following brackets

1. \((x + 2)(x - 3)\)
2. \((x+1)(x-1)\)
3. \((x+1)(x+1)\)
4. \((2x+3)(x+2)\)

Factorising brackets is the reverse of expanding them. This is much, much harder than expanding brackets. Unfortunately there is no systematic technique to factorise every quadratic equation. Some of it comes down to intuition. Let's look at an example. We'll try and factorise \(x^2 + 2x - 8\). Let's think about our goal here. We are looking for two numbers the multiply to give -8 but add to give 2. I often start by writing down all the candidate numbers that multiply to give -8. There's 1 and -8, -1 and 8, 2 and -4, -2 and 4. Do any of these pairs sum to 2? Yes! -2 and 4 sum to 2. So we can conclude that the factorisation is \((x - 2)(x + 4)\). It's easy to make a mistake when factorising equations, so always expand them again to check you have the right result. Remember, expanding brackets is easy, factorising equations is hard (this fact is used in computer science to help with security).

Factorise the following expressions

1. \(x^2 + 17x + 60\)
2. \(x^2 -5x - 36\)
3. \(x^2 - 14x + 48\)
4. \(x^2 + 2x - 24\)

What is the significance of factorising equations? Typically, with and equation we want to know its root, that is, where it crosses the x-axis. It's very easy to find the roots of a factorised equation. Consider our solution above - \((x - 2)(x + 4)\). Let's set it to 0, which will let us find the roots. In order to be 0, either of the two brackets can be 0. So we set them to 0. This gives us the equations \(x - 2 = 0\) and \(x + 4 = 0\) to solve, which should be very easy by now. The roots are \(x = 2, -4\). The graph is shown below.

Try the following on your own. Factorise the equation, and find the roots.

Sometimes, it's quite hard to factorise equations. Consider the following equation.

\[x^2 + 5x + 1 = 0\]

If you tried to solve it using the techniques you know so far, it could be quite frustrating, especially since the solutions are \(x = -\frac{5}{2} - \frac{\sqrt{21}}{2}, \frac{\sqrt{21}}{2} - \frac{5}{2}\). Not easy. Fortunately, there is a better way to solve this. We use the quadratic formula

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Where the \(a, b, c\) refer to equations in this form

\[ax^2 + bx + c = 0\]

This is one of the most important equations, so please commit it to memory. Notice that there is a \(\pm\) in the equation. This means it needs to be done twice, once for the positive and once for the negative. This reflects how quadratic equations generally have two solutions.

Let's try and solve \(x^2 + 5x + 1 = 0\) using the formula. We can see that \(a = 1\), \(b = 5\) and \(c = 1\). Plugging these in we get

\[x = \frac{-5 + \sqrt{25 - 4}}{2} = \frac{\sqrt{21}}{2} - \frac{5}{2}\]

\[x = \frac{-5 - \sqrt{25 - 4}}{2} = -\frac{\sqrt{21}}{2} - \frac{5}{2}\]

The graph for this equation looks like

There's one more thing to consider for now when it comes to quadratic equations. Consider the equation \(-x^2 + x - 1 = 0\). Let's plug it into the quadratic formula.

\[x = \frac{-1 \pm \sqrt{1 - 4}}{-2} = \frac{-1 \pm \sqrt{-3}}{-2}\]

Notice the square root? It's \(\sqrt{-3}\). The square root of a negative number doesn't exist. We would say this equation has no real roots, and the graph below of the equation should show you why. There are ways to handle these kinds of equations, but we will come back to them much later on.

Use the quadratic formula to find the roots of the following equations

1. \(x^2 + 3x - 2 = 0\)
2. \(2x^2 - 5x + 7 = 0\)
3. \(3x^2 - 4x + 1 = 0\)
4. \(4x^2 + 5x - 6 = 0\)