__Solving Quadratic Equations__

(what we've all been waiting for, right?)

So, you want to solve a quadratic equations. Well, a
quadratic equation in **standard form** is written:

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

You may recognize this as a **quadratic trinomial** set equal
to 0, which it is. One way to **solve an equation like this is by
factoring**. It may not always work, because the expression equal to 0
is not always factorable. Let's take one that is:

x^{2} + 5x + 6 = 0

Now, we know how to factor the left side of the equation, so we do so:

(x + 2)(x + 3) = 0

By a property called the **Zero Product
Rule**, if ab = 0, then a = 0, b = 0, or both = 0.

Using this, we can give a the value of x + 2, and b the value of x + 3.

Therefore, x + 2 = 0 **OR**
x + 3 = 0

**x = -2 OR x = -3**

Substitute both values back in to make sure they work:

(-2)^{2} + 5(-2) + 6 = 0

4 - 10 + 6 = 0

10 - 10 = 0 YES

(-3)^{2} + 5(-3) + 6 = 0

9 - 15 + 6 = 0

15 - 15 = 0 YES

Let's do another one:

x^{2} - 8x + 15 = 0

(x - 3)(x - 5) = 0

x - 3 = 0 **OR** x - 5 = 0

**x = 3 ****OR x = 5**

We'll do one more, that is a little harder to factor.

3x^{2} + 7x + 2 = 0

(3x^{2} + 6x) + (x + 2) = 0

3x(x + 2) + 1(x + 2) = 0

(3x + 1)(x + 2) = 0

3x + 1 = 0 **OR** x + 2 = 0

**x = -1/3 OR x = -2**

Now we will look at some special factoring patterns:

**(a + b)(a - b) = a ^{2} - b^{2
}(a + b)^{2} = a^{2} + 2ab + b^{2}**

For the next way to solve quadratic equations, we will
concentrate on using the second one. This process is called **
completing the square.**

Take x^{2} + 8x + 12 = 0**
**What we want is

x^{2} + 8x + 12 = 0

__
-12 -12
__x

Then, with this different form, we add the number needed to
complete the square on the left side, to both sides. We have to think...

In the factoring pattern, x^{2} would have
to equal a^{2}, so x
= a. 8x = 2ab. 8x
= 2xb. **Divide** each side of that equation by
2x and get 4 = b.
Because we want the constant we add to be
b^{2}, we add 16
to both sides. Does this make sense?

x^{2} + 8x + 16 = -12 + 16

(x + 4)^{2} = 4 (We factored the left side
according to the pattern, and simplified the right side.)

Now take the square root of both sides, but don't forget that on the right it
can be **positive OR negative**.

x + 4 = ±√4

x + 4 = ±2

x = -2 OR x = -6

Try another:
**x ^{2} + 6x + 11 = 0**

x

**x = a, 6x = 2xb, b = 3, b

x

(x + 3)

x + 3 =

x = ±√20 - 3

What

By the way, I'm really sorry that the lines over the radicands don't come out right (I had to underline the line above them). If you are not exactly sure what is inside and what is outside the radicand, use your common sense and bear with me (please).

Also, if a quadratic equation is ever not in standard form, change it to it by adding or subtracting terms.

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