Solving Quadratic Equation by Factoring.ppt

carling_21 58 views 14 slides Sep 17, 2024
Slide 1
Slide 1 of 14
Slide 1
1
Slide 2
2
Slide 3
3
Slide 4
4
Slide 5
5
Slide 6
6
Slide 7
7
Slide 8
8
Slide 9
9
Slide 10
10
Slide 11
11
Slide 12
12
Slide 13
13
Slide 14
14

About This Presentation

Solving Quadratic Equation by Factoring


Slide Content

Solving Equations by Factoring
Definition of Quadratic Equations
Zero-Factor Property
Strategy for Solving Quadratics

Standard Form Quadratic
Equation
Quadratic equations can be written in the form
ax
2
+ bx + c = 0
where a, b, and c are real numbers with a  0.
Standard form for a quadratic equation
is in descending order equal to zero.
BACK

Examples of Quadratic Equations
pp 1881
2

189
2
xx
25
2
y
08118
2
pp
Standard Form
0189
2
xx
025
2
y
BACK

Zero-Factor Property
If a and b are real numbers
and if ab =0, then
a = 0 or
b = 0.
BACK

Solve the equation (x + 2)(2x - 1)=0
•By the zero factor property we know...
•Since the product is equal to zero then one of the factors must be
zero.
0)2( x
2x
OR(2 1) 0x 
12x
2
1
2
2

x 2
1
x
}
2
1
,2{x
BACK

Solve the equation. Check your answers.
0)2)(5( xx
5x
OR
02x
2x
{2,5}x
05x Solution Set
BACK

Solve each equation. Check your answers.
0)35( xx
0x OR
035 x
5
3
x
}0,
5
3
{

x
0x
Solution Set
35x
5
3
5
5
x
BACK

Solving a Quadratic Equation by Factoring
Step 1 Write the equation in standard
form.
Step 2 Factor completely.
Step 3 Use the zero-factor property.
Set each factor with a variable equal
to zero.
Step 4 Solve each equation produced
in step 3.
BACK

Solve.
189
2
xx
0189
2
xx
0)3)(6( xx
}3,6{x
BACK

Solve.
07
2
xx
0)7( xx
}7,0{x
BACK

Number Of Solutions
•The degree of a polynomial
is equal to the number of
solutions.
xxx 32
23

Three solutions!!!

Example
x (x + 1)(x – 3) = 0
Set each of the three factors equal to 0.
x = 0 x + 1 = 0
x = -1
x – 3 = 0
x = 3
Solve the resulting equations.
Write the solution set.
x = {0, -1, 3}
BACK

Solve the following equations.
1.x
2
– 25 = 0
2.x
2
+ 7x – 8 = 0
3.x
2
– 12x + 36 = 0
4.c
2
– 8c = 0
x x 5 50
x x 8 10
x x 6 60
cc( )80
5,5
8,1
6
8,0

1.Get a value of zero on one side of the
equation.
2.Factor the polynomial if possible.
3.Apply the zero product property by
setting each factor equal to zero.
4.Solve for the variable.
Tags