Aim: How do we solve linear-quadratic systems algebraically?

The way we solve linear-quadratic systems algebraically is by making the two quadratic equations equal to each other.
Example:
  y=2x+8
  y=x^2+x-12
You combine both equations.
  2x+8=x^2+x-12
Then you get both equations to one side by either adding or subtracting to make the whole equation equal to zero.
  0=x^2-x-20
You do the diamond problem to find to multiples that add up to -1 and that multiplies to -20
  0=(x-5)(x+4)
  x=5, x=-4
You then plug back in both numbers for what you got for x but separately into the equations to find what y equals. Then your final answers: (5,18), (-4,0)
  

Aim: How do we solve linear-quadratic systems of equation?

The way we solve linear-quadratic systems is by combing the two equations where both Y's are equal to each other.
Example:
  y=6x
  y=4x-8
You make these to equations equal to each other:
  6x=4x-8
Then you want to get the X's by itself to find what x equals. You do this by subtracting 4x on both sides.
  2x=-8
Then you divide 2 on both sides to get what X equals.
  x=-4
You then plug in what you got for X in either equations.
y=6(-4)
y=-24
That's your solution: (-4,-24)

Aim: How do we solve quadratic equations by taking square roots?

If b^2-4ac is a perfect square, then the two roots are rational.
If b^2-4ac isn't a perfect square, then the two roots are irrational.
Examples:

y=x^2+6x+8

a= x^2
b=6
c=8
b^2-4ac= 6^2-4(1)(8)
36-4(8)
36-32=4
Rational b/c 4 is a perfect square

y=x^2=3x-1
a=1
b=3
c=-1
b^2-4ac= 3^2-4(1)(-1)
9-4(-1)
9+4=13
Irrational b/c 13 isn't a perfect square

Aim: How do we use the discriminant to find the number of solutions to a quadratic equation?

The discriminant is: b^2-4ac
If b^2-4ac is greater than 0, then you'll get two solution's.
If b^2-4ac is equal to 0, then you'll get one solution.
If b^2-4ac is less than 0, then you'll get no solutions.

When you have a quadratic equation and your using the discriminant, you first find what a, b, and c equal.
Quadratic equation example: x^2+6x+9=0
A equals: x^2 of a quadratic equation
B equals: 6x
C equals: 9
You then use the discriminant: b^2-4ac, to find how many solutions the quadratic equation has.

Aim: How do we divide radicals?

When dividing radical expressions, as long as the roots are the same, we can divide the radicals using the following property. You can divide the radicala first or simplify first then divide. Either way you choose to work the problems the results will be the same.
 http://infinity.cos.edu/algebra/ProblemsSolved/Chapter%2009/Chapter%209.4_Multiplication%20and%20Division%20of%20Radicals.pdf

Aim: How do we multiply and divide radical expressions?

The way we multiply and divide radical expressions is by compining the factor and radical together. Then simplifying the radical when multiplying. When dividing a radical you do the same with multplying but then simplfy the radicals.

Aim: How do we simplify radicals?

The way we can simply radicals is finding the multipls of the factor and taking out any pairs of numbers outside of the radical.
Example:

     The radical 52 has the factors of 2x2x13 which equal 52. Since 2 has a pair of 2 which makes 4, you bring the 2^2 out of the radical and leave 13 inside the radical. Since the square root of 2 is 2, you take out the power of 2 and leave it as is for the final answer.

Aim: How do we solve quadratic equations by factoring?

When the product of two factors are equal to zero, at least one of the two factors must equal zero
Example:
   If 10(x+17)=0, then 10 or x+17 must be zero
     x+17=0
     x=-17
  If (x+9)(x-13)=0, then x+17 or x-13 must equal to zero
     x+9=0, x=-9
     x-13=0, x=13

Aim: How do we solve rational proportions?

How you solve rational proportions is by cross multiplying two fractions that are equal to each other to find the value of a variable. After cross multiplying the two fractions, your left with two numbers equaling each other, you then divide the number with the variable to both sides to find with the variable is equal to. 

Aim: How do we add and subtract rational expressions?

The way you add and subtract rational expressions is by trying to get all the denominators of the fractions to be the same. Then cancel out all the denominators when they're the same and solve the equation for the variable that's missing.
    

Aim: How do we solve diamond problems?

One way you can solve diamond problems is by putting the lowest number in addition and the highest in the multiplication. Find to factors that both numbers have in common and put both numbers in to equations; bring all the signs down with it.

Aim: How do we multiply rational expressions?

There are two methods when multiplying and solving for rational expressions. In some problems when multiplying rational expressions, there are two steps that you'll need to do in order to solve the problem.

     When multiplying rational expressions, you want the terms to be in simplest form. In order to do this, you'll have to use the diamond system and cross out all terms that are the same. Then you put the problem together as your final fantasy.

Aim: How do we multiply bionomials?

One method in multiplying bionomials is using the FOIL system which stands for first, inner, outter, last; and combining any like terms.
Example:  (2x+5)(x-2)
When using the FOIL method, you first distribute the 2x from the first equation in parenthesis to the second equation in parenthesis with the distrubutive property of multiplication and bring down the sighs:
(2x)(x)+(2x)(-2)= 2x^2-4x
Then from the first numbers in the parenthesis [(2x+5)] you use the same method, distrubtuive property of multiplication and mulitply 5 this time into the second equation and again, bring down the sighs:
(5)(x)+(5)(-2)=5x-10
You then combine the equation together: 2x^2-4x+5x-10
And combine any like terms as to being your final answer: 2x^2+x-10

Aim: How do we multiply polynomials?

One way we can multiply polynomials is by using the distrubutive property of multiplication and then combining any like terms.
Example: 2x(3x+1)
You distrubute the 2x on the out side of the parenthesis to 3x and 1 and bring down the sign
It would look like this : (2x)(3x)+(2x)(1)
Multiply  : 6x^2+2x