### Basic Overview of Quadratic Equations

In the article on Factorising we touched briefly on how to factorise Quadratic Equations.

We’ll now go a bit more in-depth and talk a bit more about solving Quadratic Equations using various techniques.

Quadratic Equations are one of the most useful equations in Mathematics. Although they may seem confusing at the moment, quadratic equations are vital to pretty much every branch of Mathematics, from Fluid Mechanics to Population Studies.

Let’s start from the basics. Quadratic Equations are equations in which one of the terms involves an $x^2$. They may be of the form:

1) $ax^2 + bx + c$

or

2) $ax^2 + bx$

or

3) $ax^2 + c$

Here the letters a, b and c are just used to represent numbers.

Whatever form your quadratic equation is in, the first thing you need to do in solving it is to get a zero on one side of the brackets. It’s also a good idea, if you have a minus number in front of your $x^2$ term, to multiply everything by -1, as this will make your calculations easier. Here’s a couple of examples of type 2) and 3).

Example 1

$2x^2 = 5x$

First we move everything over to the left hand side (LHS) of the equation:

$2x^2 - 5x = 0$

Now notice that both terms have a common factor of $x$, so we factorise like so:

$x(2x - 5) = 0$

You should know that if two things multiplied together are equal to zero, then one them is going to be zero.

Hence we have either $x = 0$, or $2x-5=0$, so:

$x = 0$, or $x=2.5$

Example 2

$-x^2 = -7$

Because this only contains one $x$ term and one number, this is pretty simple to solve.

First we multiply both sides by -1 to get $x^2 = 7$, and then simple take the square root of both sides.

$x = \pm \sqrt{7}$

The $\pm \sqrt{7}$ simply means “plus or minus root 7”, this is because when you take the square root, you always get a positive and negative value.

Warning – It’s important to remember that you can’t take the square root of a negative number!

Now you’ve learn now to solve the two easier types of quadratic equation, now let’s move on to the hardest.

For quadratic equations of the form $ax^2 + bx + c = 0$, there are generally 3 methods we can use to solve them:

#### Factorising

I won’t go into detail here about how to factorise quadratic equations, as this is discussed in the article on factorising. Let’s do an example of factorising a quadratic equation to solve it.

Example 3 – Solve the equation $2x^2 - 4x + 5 = x^2 - 8x + 2$ by factorising.

First we need to rearrange our equation to get one side equal to zero. It shouldn’t be hard to see that the above equation can be rearranged to $x^2 + 4x + 3 = 0$. Now by using the methods discussed in the article on factorising, we can write our equation as $(x + 1)( x + 3) = 0$.

If two things multiplied together are equal to zero, then one of our brackets must be equal to zero. Hence we have our solution, $x = -1$ or $x = -3$

#### Completing the square

Completing the Square is a method used to solve quadratic equations that can’t easily be factorised or solved by other methods. At first it can seem a little complicated but after a bit of practice you’ll be an expert in to time.

Example 4 – Solve the equation $2x^2 - 2x - 2 = 0$ by completing the square.

Before we start anything else, when completing the square, we need to make sure that the number in front of the $x^2$ is one, so we divide our equation by 2 to get $x^2 - x - 1 = 0$

The next thing we do is to take number not involving $x$ and ‘move it over’ to the other side.

$x^2 - x = 1$

Now the next step is a little confusing, but what we do is take $x$ plus half the coefficient of our original $x$, and square them.

$(x - \frac{1}{2})^2 = 1$

Now the more observant readers amongst you may have noticed that this is not quite right. If we expand the brackets, $(x - \frac{1}{2})^2 = x^2 - x + \frac{1}{4}$. To compensate for this extra $\frac{1}{4}$, we place a $-\frac{1}{4}$ in our equation, giving:

$(x - \frac{1}{2})^2 - \frac{1}{4} = 1$

Still with us? Good. Now what we do is to move all of the numbers over to the left hand side, giving:

$(x - \frac{1}{2})^2 = \frac{1}{4} + 1 = \frac{5}{4}$

Now we take square roots of our equation

$x - \frac{1}{2} = \pm \frac{\sqrt{5}}{2}$, (because $\sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{\sqrt{4}} = \frac{\sqrt{5}}{2}$, and remembering that taking the square root leaves you with a positive and minus value).

Now we simple add $\frac{1}{2}$ to both sides of our equation and we have our answer.

$x = \frac{1}{2} + \frac{\sqrt{5}}{2}$ or $x = \frac{1}{2} - \frac{\sqrt{5}}{2}$

Now this method is a little tricky to get to grips with. If you’re having problems then the best thing will be to ask your teacher for some extra work sheets on completing the square and practice on those.

For a quadratic equation in the form $ax^2 + bx + c = 0$, the following formula give you the solutions:

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

Example 5 – Solve the equation $x^2 - x - 1 = 0$ using the quadratic formula.

The first thing we do is to write down the values of $a,b,c$. We have $a=1$, $b=-1$ and $c=-1$.

Now we use our quadratic formula to find the values of $x$

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

$= \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)}$

$= \frac{1 \pm \sqrt{(1 - (-4)}}{2}$

$= \frac{1 \pm \sqrt{5}}{2}$

Which you will notice is the same result that we got by Completing the Square.

Warning – You will remember that you can only take the square root of a positive number. In some cases, quadratic equations have no solutions, this is when you end up with a square root of a minus number in your solution.

If you look back at the quadratic formula, this only occurs with b^2 – 4ac < 0[/latex]. $b^2 - 4ac$ is called the discriminant of the quadratic, and can be used to tell things about the roots (solutions) of the quadratic.

If the discriminant is greater than zero, then your equation will have 2 distinct (different) roots.

If the discriminant is equal to zero, then your equation will have 1 repeated root.

If the discriminant is less than zero, you end up with the square root of a minus number, which is impossible, so your equation has no solution.

You don’t have to remember the Quadratic Formula as it is given to you in the formula sheet in your exam. For those of you who are interested, there’s an article on how the quadratic formula is derived.

So there you have it. Which method to use depends on the question. Sometimes they will ask you to use a certain method, sometimes it comes down to practice which one to use.

Remember the only way to be good at anything in maths, especially quadratic equations, is lots of practice!