Aim: how do we solve quadratic inequalities ? 

To solve quadratic inequality, first you can either graph or factor the inequality to see where the graph hits the x-axis. Then test different points to see which points make the inequality true. 

For example,
y< x^2-x-12  
First, factor or graph it to find where the parabola hits the x-axis 


y< x^2-x-12  
y = x^2-x-12  

Zero Product Property:
0= x^2-x-12  
0 =(x+3)(x-4)

x+3=0                         x-4=0
  -3  -3                           +4 +4   
x= -3                              x=4

Now, to find out where the points will make the inequality true, pick test points from both inside and outside the graph to test.


Inside:
(0,0)
0< (0)^2-(0)-12
0< 0-0-12
0< -12              FALSE


Outside: 
(-4,-8)
-8< (-4)^2-(-4)-12
-8< 16+4-12
-8< 20-12
-8< 8                   TRUE


Finally, the solution is: x<-2 or x>4 


Now that you know the points outside the parabola makes the graph true, you shade the outside of the parabola.

How do we use complex conjugate to divide imaginary numbers?

How do we use complex conjugate to divide imaginary numbers?

To divide imaginary numbers you write the quotient as a fraction and divide the numerator and the denominator  by the conjugate of the denominator. 

The complex conjugate of (a+bi) is (a-bi)

* Note you just change the sign in the middle. 

Example:

(1-2i)/ (3+i)

-First, find the complex conjugate of the denominator 

-Complex conjugate of (3+i) is (3-i)

-Next multiply both numerator and denominator by (3-i)

-If you foil (1-2i) and (3-i) you get 1-7i

-If you foil (3+i) and (3-i) you get 10

-Your final; answer is 1-7i/10

    

*Note i x i= i^2= -1

How to foil?

Citation (images):- http://literacy.kent.edu/Oasis/Resc/Educ/algebra.html

                          -http://www.virtualnerd.com/algebra-2/quadratics/imaginary-complex-numbers.php