Section 3.2 Solving Quadratic Equations
Today:
Return midterms and review.
3.2 Quadratic Equations
Announcements
 Homework 3.2A due Fri

Recall:
solving a linear equation:
25 x  20  620
25 x  600
x  30
(highest power of x is 1)
A quadratic equation:
25 x 2  20 x  620 (highest power
of x is 2)
We will learn 4 ways to solve quadratic equations:
(1) by factoring and applying the principle of zero products (sometimes possible)
(2) by using the principle of square roots (sometimes possible)
(3) by completing the square
(4) using the quadratic formula (always possible)
1. By Factoring and Applying Principle of Zero Products
Factoring: R.4
Notice that (x+A)(x+B) = x2 + (A+B)x+AB.
Example: (x+3)(x-4) = x2 + 3x – 4x – 12 = x2-x-12
where 3×-4 = -12 and 3+-4 = -1
Problem: Factor x2+7x-30
(1) p+q = b
-1+30=29
-2+15=13
-3+10=7
-5+6=1
Rule for factoring:
x2+bx+c will factor into
(x+p)(x+q) IF
(1) p+q = b
(2) p*q = c
(2) p*q = c
-1×30
-2×15
-3×10
-5×6
Problem: Solve x2+7x= 30
x2+7x = 30
The principle of zero
2
x +7x-30 = 0
products: If ab=0 is true if
(x-3)(x+10) = 0  x-3=0 OR a=0 or b=0.
x+10=0
2. By Using the Principle of Square Roots
6x2 = 36
x2-16=0
x2 = 6
x2 = 16
x=± 6
x= ± 16 = ±4
Another example:
18x+9x2 = 0
9x(2+x)=0
x2+16=0
x2 = -16
x= ±  16 = ±4i