3. Solve for *x* and check

2*x*^{2} - *x* - 10 = 0

This is a quadratic (degree 2) equation. The steps are not all the same as solving first degree equations. We are responsible for three different methods for solving quadratic equations. The first is

- Clear denominators. Multiply both sides by a common denominator.
- Simplify. Remove parentheses and combine like terms.
- Transpose all terms to one side and leave a 0 on the other.
- Factor.
- Set the factors = 0
- Solve.
- Check.

Amazingly, you will find that factoring it the quickest and easiest method for solving a quadratic equation. The trouble is that when you get to step 4, if might not factor. In that case we can use the second method

- Clear denominators. Multiply both sides by a common denominator.
- Simplify. Remove parentheses and combine like terms.
- Transpose known terms to one side and unknonwn terms to the other.
- Combine.
- Divide both sides by the coefficient of the square term.
- Add the square of half the coefficient of the first degree term to both sides.
- Simplify to get a perfect square on one side and a number on the other.
- Take square roots of both sides
- Transpose
- Check.

Completing the square is very ancient. The Babylonians knew it twenty-five hundred years ago, and it may even be older than that. As you can see, it is more complicated than factoring, but it will always work, where factoring may not. The third method, the quadratic formula, is not as ancient. It uses algebra, and algebra is credited to the Arab in the 8th century AD. As soon as the Arabs came up with the idea of arithmetic with unknowns, they were able to use completing the square to derive the quadratic formula.

You will always try factoring because if it works it will be quicker and easier. If it doesn't work you will find it out after step 4 in the factoring method. At this point, your euqation will look like

*ax*^{2} + *bx* + *c* = 0

We apply completing the square to this. We have cleared denominators and simplified. We are ready to transpose known terms to the other side.

*ax*^{2} + *bx* = -*c*

Now we're ready to divide both sides by *a.*

The next step, "6. Add the square of half the coefficient of the first degree term to both sides." is probably the most complicated of the steps. We skip straight to the next step. Recall

(*x* + *n*)^{2} = *x*^{2} + 2*nx* + *n*^{2}

so we seze that the two terms on the left could be the first two terms in a perfect square. We would just need to have

2*n* = *b*/*a*

*i.e.*,

*n* = *b*/2*a*

Note

So to get from where we were to our perfect square, we would need to add b2/4a2 to the left side of the equation. We can add whatever we want to one side of an equation so long as we add the same thing to the other side.

Before we simplify the left side to get a perfect square, we should find common denominators and all the terms on the right.

Thus

This acheives the very desireable outcome of getting all of the *x*'s together in the same place. Once that happens, all we have to do is to undo all of the arithmetic which has happened to *x* in the reverse order to the way it was applied. First take square roots.

Then all we really have to do is to transpose the b/2a to the other side of the equation. At the same time, let us siimplify the radical on the right. The bottom under the radical is a perfect square.

Note that we have common denominators on the right. This will give us the quadratic formula in the form which is most commonly used in America.

Click here for a worked example using factoring.

Click here for a worked example using completing the square.

Click here for a worked example using the quadratic formuula.