Download Quadratic Equations Section 1.3

Solving Quadratic Equations
Section 1.3
What is a Quadratic Equation?
A quadratic equation in x is an equation
that can be written in the standard form:
ax² + 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 x² - 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 3t² + 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:
3t² + 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.
2 x  3x  0
x2 x  3  0
2
x0
2x  3  0
3
x
2
Solve by the Square Root Method.
If the quadratic has the form ax² + 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 x² = c, then x = ±c.
Note: The variable squared must be isolated first (coefficient
equal to 1).
Example 1:
Solve by the Square Root Method:
2x² - 32 = 0
2x² = 32
x² = 16
x 2 = 16
x=±4
Example 2:
Solve by the Square Root Method.
5x² + 10 = 0
5x² = -10
x² = -2
x = ± 2
x = ±i 2
Example 3:
Solve by the Square Root Method.
(x – 3)² = 25
x–3=±5
x – 3 = 5 or x – 3 = -5
x=8
x = -2
Solve by the Square Root Method
3x  82
 12
3x  8   12
3x  8  2 3
3x  8  2 3
3x  8  2 3
3x  8  2 3
8 2 3
x
3
8 2 3
x
3
Solve by Completing the Square.




Words
Express the quadratic
equation in the following
form.
Divide b by2 and square the
result, then add the square
to both sides.
Write the left side of the
equation as a perfect
square.
Solve by using the square
root method.

Math
x² + bx = c

b
b
x² + bx + ( )² = c + ( )²
2
2

(x + b )² = c + ( b )²
2
2
Example 1:
Solve by Completing the Square.
x² + 8x – 3 = 0
x² + 8x = 3
x² + 8x + (4)² = 3 + (4)²
x² + 8x + 16 = 3 + 16
(x + 4)² = 19
x + 4 = ± 19
x = -4 ± 19

Add three to both sides.

Add ( b )² which is (4)² 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
your two solutions.
Example 2:
Solve by Completing the Square when the
Leading Coefficient is not equal to 1.
2x² - 4x + 3 = 0
x² - 2x +
3
=0
2
3
x² - 2x + ___ = + ____
2

Divide by the leading
coefficient.

Continue to solve using the
completing the square method.

Simplify radical.
x² - 2x + 1 = 3 + 1
2
5
(x – 1)² =
2
5
2
x–1=±
x=1±
10
2
Quadratic Formula
If a quadratic can’t be factored, you must
use the quadratic formula.
If ax² + bx + c = 0, then the solution is:
 b  b 2  4ac
x
2a
a=1
Solve
x 2  4x  1  0
 b  b 2  4ac
x
2a
x
4
 42  41 1
2 1
4  16  4
x
2
4  20
x
2
b = -4
c = -1
42 5
x
2
x 2 5
Solve
4n 2  6n  9
4n 2  6n  9  0
 b  b 2  4ac
n
2a
 6  62  44 9
n
24
 6  36  144
n
8
 6  180
n
8
66 5
n
8
33 5
n
4
Solve
x2  8  4 x
x2  4x  8  0
 b  b 2  4ac
x
2a
x
4
 42  418
2 1
4  16  32
x
2
4   16
x
2
4  4i
x
2
x  2  2i
Discriminant
 b  b 2  4ac
x
2a
The term inside the radical b² - 4ac is called the
discriminant.
The discriminant gives important information about the
corresponding solutions or answers of ax² + bx + c = 0,
where a,b, and c are real numbers.
b² - 4ac
b² - 4ac > 0
b² - 4ac = 0
b² - 4ac < 0
Solutions
Tell what kind of solution to expect
x 2  28 x  198  0
b 2  4ac  282  41198
 784  792
 8