- Topics At a Glance
- Squares
- Simplification of Radical Terms
- Multiplication
- Division
- Radicals in the Denominator
- Radical Arithmetic
- Addition and Subtraction
- Multiplication
- Division
**Quadratic Equations**- Taking Square Roots
- Completing the Square
**Quadratic Formula**- Solving Radical Equations
- The Pythagorean Theorem
- Word Problems
- In the Real World
- I Like Abstract Stuff; Why Should I Care?
- How to Solve a Math Problem
- Yes, This Really Is a Square

Let's kick things off with a...

What, you were expecting a party?

Solve the quadratic equation *ax*^{2} + *bx* + *c* = 0 by completing the square.

Here we know a, b, *c* are numbers, but we don't know what any of them are. We do know that *a* can't be 0, or we wouldn't have a quadratic equation. We have a sneaking suspicion that *b* is 17, but that's only based on a dream we had last night, so we should probably do the math to be on the safe side.

First, we divide both sides by *a* to find that

.

Then we subtract from both sides to get it out of the way:

.

Next we take (the coefficient on *x*), divide by 2, and square to find . It's not as nice-looking as what we've had in the past, but we'll go with it. This is the number we add to both sides:

.

The left-hand side of the equation can now be written as a square:

.

Since

we can add the numbers on the right-hand side of the equation:

It's a good feeling to look at such an ugly conglomeration of numbers, variables, fractions and parentheses and know that you can make some sense out of it, eh? Not that you want to print it out and hang it on your bedroom wall, but still. We think it's nice.

Now, writing the left-hand side of the equation as a square and the right-hand side of the equation in its new form, we need to solve the equation

.

Since we have the square of a degree 1 polynomial on the left and an admittedly messy number on the right, we know where to go from here. We take a square root to find

.

We can simplify the right-hand side:

.

Now, we need to solve the two equations

.

We find, respectively, the solutions

and

These solutions are usually written together as

.

and called the **quadratic formula**.

You may have already heard of this one. It's famous. It used to have its own talk show, and it was voted one of People's Sexiest Formulas Alive in 300 BCE.

Okay, maybe it hasn't achieved *that* kind of fame, but it is well known. The quadratic formula is another way we can find solutions to quadratic equations. When given a quadratic equation, we figure out which numbers correspond to a,b, and *c*, then plug them into the quadratic formula to find our answers. It's like one of those factory machines where you throw in various ingredients, and then out comes a gloriously wonderful Oreo, or Twinkie, or wax lips. Yes, we went with "wax lips."

As complicated as it is to arrive at and understand the quadratic formula, it is fortunately one of those things that's best to memorize and use. It's unlikely that you'll be asked in school to explain where the quadratic formula comes from, but it's reassuring to know that, like babies, it does come from somewhere.

Try to make it feel at home, won't you?

Solve: *x*^{2} + 5*x* – 7 = 0.

The coefficient on *x*^{2} is 1, the coefficient on *x* is 5, and the constant term is -7, so we have

All we do is plug these numbers into the quadratic formula,

and we have our solution in no time:

Since 53 is a prime number, we can't simplify any more. The two solutions to the equation are

.

**Be Careful:** In order to use the quadratic formula, you need an equation of the form

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

Because we must have zero on one side of the equation, you can't go quadratic equation-happy and start plugging everything under the sun into it. By the way, if you ever go overseas, make sure you pack your European quadratic plug adapter.

A student, the same hopeless daydreamer who aspires to one day be a member of the National Audubon Society, was asked to use the quadratic formula to solve the equation *x*^{2} + 3*x* – 6 = 7.

She wrote down the following work. What did she do wrong?

The original equation did not have zero on one side of the equation. D'oh! The student should have first subtracted 7 from each side, then solved the equation *x*^{2} + 3*x* – 13 = 0.

This is starting to become a problem. Her teacher might want to consider drawing the blinds.

A different student, one who is not interested in birds so much as he is captivated by the incoming texts on the cell phone he is discreetly concealing in one of his sleeves, was asked to use the quadratic formula to solve the equation 4*x*^{2} + 15*x* – 9 = 0.

He wrote down the following work. What did he do wrong?

In the original equation, *c* was -9, not +9. He might have noticed that had he not been so busy lol'ing and jk'ing.

4*x*^{2} + 15*x*– 9 = 0

The student's first two lines should have been

Sometimes a quadratic equation has no solutions. The quadratic equation says

.

Since we can't take square roots of negative numbers when we're looking for real number solutions, this only works if *b*^{2} – 4*ac* is at least zero. The expression *b*^{2} – 4*ac* is called the **discriminant** of the equation

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

The quantity *b*^{2} – 4*ac* discriminates between those equations that have real number solutions and those that don't.

Example 1

Solve: 3 |

Example 2

Solve: |

Example 3

Solve: |

Exercise 1

For the following equation, solve using the quadratic formula or state that there are no real number solutions: 10*x*^{2} + 9*x* = 0.

Exercise 2

For the following equation, solve using the quadratic formula or state that there are no real number solutions: 2*x*^{2} + 11*x* = 10.

Exercise 3

For the following equation, solve using the quadratic formula or state that there are no real number solutions: 5*x*^{2} – 3*x* – 1 = 0.

Exercise 4

For the following equation, solve using the quadratic formula or state that there are no real number solutions: *x*^{2} – 2*x* + 3 = 0.

Exercise 5

For the following equation, solve using the quadratic formula or state that there are no real number solutions: *x*^{2} – 4*x* – 7 = 8.