1. Solving Quadratic Equations
Pre-Calc Lesson 1.6
Solving Quadratic Equations
Quadratic Equation - Any equation that can be
written in ax2 + bx + c = 0 form.
**Three methods for solving quadratic equations:
1. Factoring
2. Completing the square! 
3. Quadratic formula! 
Example 1: Solve by factoring
(3x – 2)(x + 4) = - 11
Since this is not = 0, we must first expand by using ‘foil’
3x2 + 12x – 2x – 8 = - 11
3x2 + 10x + 3 = 0
Factor: (3x + 1)(x + 3) = 0
3x + 1 = 0 and x + 3 = 0
x = -⅓ & x = - 3

2. Example 2: Solve by Completing the Square! 2x2 – 12x – 7 = 0 ???
1st divide all by 2 so coefficient of x2 is 1
x2 – 6x – 7/2 = 0
Now isolate the ‘x’ terms  add 7/2 to both sides
x2 - 6x = 7/2
Now take ½ of ‘b’, then ‘square it and add that result to both sides.
-6/2 = (-3)2 = 9
x2 – 6x + 9 = 7/2 + 9
(x – 3)2 = 25/2
take the square root of both sides
√(x – 3)2 = √25/2
x – 3 = + 5/√2 Now rationalize this
x – 3 = + 5√2 2
so  x = 3 + 5√2

3. Example 3: Solve by using the Quadratic Formula!!! 2x2 + 7 = 4x
1st get in ax2 + bx + c = 0 order
2x2 – 4x + 7 = 0
a = 2, b = - 4, and c = 7
Thus: x = - (-4) + √(-4)2 – 4(2)(7)
2(2)
x = 4 + √16 – 56 4
x = 4 + √-40
x = i√10
now factor a ‘2’ from all three terms and get:
x = 2 + i√10 2

4. Discriminant. In the Quadratic Formula we have the expression
under the radical b2 – 4ac This is called the
Discriminant.
Is it so appropriately named because it tells us something
special (or discriminates) about our roots!(x-intercepts)
It can tell us many things about our solutions. (roots, zeros)
For Instance, if:
b2 – 4ac < there will exist 2 complex conjugate roots
b2 – 4ac = There will only exist 1 real root – called
a double root
3. b2 – 4ac > There will exist ‘2’ ‘distinct’ real roots

5. Some Helpful Hints: Quadratic Formula!! 
Sometimes when sitting by oneself doing one’s Pre-Calc
Homework, one may think, “Self, which method should
I use to solve this here quadratic equation?” 
Some Helpful Hints:
If a,b, & c are integers, and if b2 – 4ac is a perfect
square, then factor the sucker!
If the equation has the form: x2 + (even #)x + c = 0,
then solve by completing the square!!
If ‘niether’ of those two cases exist, then use the
Quadratic Formula!! 

6. Two special circumstance to look for:
(Possibly losing a root! )
Take the equation: x(x – 1) = 3(x – 1)2
One approach might be:
Divide both sides by (x – 1)  4x = 3(x – 1)
4x = 3x – 3
x = - 3 !!
This will only give us one ‘root’  which may not be
wrong, but if we check the answer x = 1, which comes
from the (x – 1) factor we divided out in the original
problem we would see that it makes the equation true
also!
Therefore never discard a factor with a variable in it!

7. Case 2: Gaining a root! Check for Extraneous roots!
Example: x x – 2 = 8 – 4x
x – x x2 – 4
1st – find the least common denominator of all the ‘fractions’ = (x – 2)(x + 2)
Now multiply each and every term by the LCD!
(x – 2)(x + 2) [x x – 2] = (x – 2)(x + 2)[ – 4x ]
x – x (x – 2)(x + 2)
(x + 2)(x + 2) + (x – 2)(x – 2) = 8 – 4x
x2 + 4x x2 – 4x = 8 – 4x  4x’s cancel
2x2 + 8 = 8 – 4x  8’s xancel
2x2 + 4x =  factor out a ‘2x’

8. 2x(x + 2) =  soooooo
2x = 0 and x + 2 = 0
x = 0 and x = butttt 
if we try to insert and check x = -2 back in the original
equation we will get an undefined expression
 sew ???
x = - 2 is called an extraneous solution and we
must discard it sooooo 
x = 0 is our only solution!
Hw Pg ????? We will get in class on Monday!