 ```Solving Quadratic Equations
Section 1.3
A quadratic equation in x is an equation
that can be written in the standard form:
ax&sup2; + bx + c = 0
Where a,b,and c are real numbers and
a ≠ 0.
Solving a Quadratic Equation by Factoring.
The factoring method applies the zero product property:
Words: If a product is zero, then at least one of its
factors has to be zero.
Math:
If (B)(C)=0, then B=0 or C=0 or both.
Recap of steps for how to solve by
Factoring




Set equal to 0
Factor
Set each factor equal to 0 (keep the squared
term positive)
Solve each equation (be careful when
determining solutions, some may be
imaginary numbers)
Example 1
Solve x&sup2; - 12x + 35 = 0 by factoring.

Factor:

(x – 7)(x - 5) = 0

Set each factor equal to zero
by the zero product property.

(x – 7)=0

Solve each equation to find
solutions.

x = 7 or x = 5

The solution set is:

{ 5, 7 }
(x – 5)=0
Example 2
Solve 3t&sup2; + 10t + 6 = -2 by factoring.

Check equation to make sure it is in standard form before
solving. Is it?

It is not, so set equation equal to zero first:
3t&sup2; + 10t + 8 = 0

Now factor and solve.
(3t + 4)(t + 2) = 0
3t + 4 = 0
t=
4
3
t +2 = 0
t = -2
Solve by factoring.
2x  3x  0
x 2 x  3   0
2
x0
2x  3  0
x
3
2
Solve by the Square Root Method.
If the quadratic has the form ax&sup2; + c = 0, where a ≠ 0, then we
could use the square root method to solve.
Words: If an expression squared is equal to a constant, then that
expression is equal to the positive or negative square root of
the constant.
Math: If x&sup2; = c, then x = &plusmn;c.
Note: The variable squared must be isolated first (coefficient
equal to 1).
Example 1:
Solve by the Square Root Method:
2x&sup2; - 32 = 0
2x&sup2; = 32
x&sup2; = 16
2
x = 16
x=&plusmn;4
Example 2:
Solve by the Square Root Method.
5x&sup2; + 10 = 0
5x&sup2; = -10
x&sup2; = -2
x=&plusmn; 2
x = &plusmn;i 2
Example 3:
Solve by the Square Root Method.
(x – 3)&sup2; = 25
x–3=&plusmn;5
x – 3 = 5 or x – 3 = -5
x=8
x = -2
Solve by the Square Root Method
3 x  8  2
 12
3 x  8   12
3x  8  2 3
3 x  8  2 3
3 x  8  2 3
3 x  8  2 3
82 3
82 3
x
3
x
3
Solve by Completing the Square.




Words
equation in the following
form.
Divide b by2 and square the
to both sides.
Write the left side of the
equation as a perfect
square.
Solve by using the square
root method.



Math
x&sup2; + bx = c
x&sup2; + bx + (
b
2
)&sup2; = c + (
(x + b )&sup2; = c + ( b )&sup2;
2
2
b
2
)&sup2;
Example 1:
Solve by Completing the Square.
x&sup2; + 8x – 3 = 0
x&sup2; + 8x = 3
x&sup2; + 8x + (4)&sup2; = 3 + (4)&sup2;
x&sup2; + 8x + 16 = 3 + 16
(x + 4)&sup2; = 19
x + 4 = &plusmn; 19
x = -4 &plusmn; 19


Add ( b )&sup2; which is (4)&sup2; to both
sides. 2

Write the left side as a perfect
square and simplify the right side.
Apply the square root method to
solve.
Subtract 4 from both sides to get


Example 2:
Solve by Completing the Square when the
Leading Coefficient is not equal to 1.
2x&sup2; - 4x + 3 = 0
x&sup2; - 2x +
3

coefficient.

Continue to solve using the
completing the square method.

=0
2
x&sup2; - 2x + ___ =
3
+ ____
2
x&sup2; - 2x + 1 = 3 + 1
(x – 1)&sup2; =
x–1=&plusmn;
x=1&plusmn;
5
2
5
2
10
2
2
If a quadratic can’t be factored, you must
If ax&sup2; + bx + c = 0, then the solution is:
x
b
b  4 ac
2
2a
a=1
b = -4
Solve
x
x  4x 1  0
2
b
b  4 ac
2
2a
x
x
4
 4 
x
2
42 5
 4 1  1
2
2 1
4
16  4
2
x
c = -1
4
20
2
x2
5
Solve
4n  6n  9
2
4n  6n  9  0
2
n
b
b  4 ac
2
n
6
180
8
2a
n
n
6
6  4  4   9 
2
n
66 5
8
24
6
36  144
8
n
33 5
4
x  8  4x
2
Solve
x  4x  8  0
2
x
b
x
4
2
b  4 ac
2
2a
 16
x
4  4i
2
x
4
x
 4 2  4 18 
2 1
4
16  32
2
x  2  2i
Discriminant
x
b
b  4 ac
2
2a
The term inside the radical b&sup2; - 4ac is called the
discriminant.
The discriminant gives important information about the
corresponding solutions or answers of ax&sup2; + bx + c = 0,
where a,b, and c are real numbers.
b&sup2; - 4ac
b&sup2; - 4ac &gt; 0
b&sup2; - 4ac = 0
b&sup2; - 4ac &lt; 0
Solutions
Tell what kind of solution to expect
x  28 x  198  0
2
b  4 ac  28  4 1198 
2
2
 784  792
 8
```