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.