Download Solving Quadratic Equation using Factoring

MTH 10905
Algebra
Solving Quadratic
Equation using
Factoring
Chapter 5 Section 6
Recognize Quadratic Equation
 Quadratic Equation in Standard Form
ax2 + bx + c = 0
a, b, and c are real numbers a ≠ 0
 Examples of a quadratic equation in standard form:
x2 + 4x – 12 = 0
2x2 – 5x = 0
same as
2x2 – 5x + 0 = 0
3x2 – 2 = 0
same as
3x2 + 0x – 2 = 0
Zero Factor Property
 ab = c
What do you know about a and b if their product is 6?
ab = 6
6
6
and
b=
a
b
 What do I know about a and b if their product is 0?
a=
ab = 0
0
a=
b
a=0
and
b=
0
a
b=0
Zero Factor Property
 Zero Factor Property
a • b = 0, then a = 0 or b = 0
 To solve a Quadratic Equation by factoring we
use the Zero Factor Property.
 We know that is we multiply by 0 the product
will be 0.
 If a product equals 0 then one of the factors
has to be 0.
Solve the Equation
Example:
(x + 8) (x + 2) = 0
x + 8 = 0 and x + 2 = 0
x = -8
x = -2
To write (x+8)(x+2) in standard form you must multiply using FOIL.
x2 +10x +16 = 0
Does the above solutions solve the equation? YES
(-8)2 +10(-8) +16 = 0
64 – 80 + 16 = 0
-16 + 16 = 0
(-2)2 +10(-2) +16 = 0
4 – 20 + 16 = 0
-16 + 16 = 0
Solve the Equation
Example:
(2x + 5) (3x – 7) = 0
2x + 5 = 0 and 3x – 7 = 0
2x = -5
3x = 7
5
7
x
x
2
3
Check 1st solution: (2x + 5) (3x – 7) = 0
[(2)(-5/2) + 5] [(3)(-5/2) – 7] = 0
[-10/2 + 5] [-7.5 – 7] = 0
[-5 + 5] [-14.5] = 0
[0] [-14.5] = 0
0=0
Solve the Equation
Example:
(2x + 5) (3x – 7) = 0
2x + 5 = 0 and 3x – 7 = 0
2x = -5
3x = 7
5
7
x
x
2
3
Check 2nd solution: (2x + 5) (3x – 7) = 0
[(2)(7/3) + 5] [(3)(7/3) – 7] = 0
[14/3 + 5] [21/3 – 7] = 0
[29/3] [7 – 7] = 0
[29/3] [0] = 0
0=0
Solving Quadratic Equation
using Factoring
1. Write the equation in standard form with the
squared term having a positive coefficient.
This will result in one side of the equation
being 0.
2. Factor the non-zero side.
3. Set each factor containing a variable equal to
zero (0) and solve the equation.
4. Check each solution in the original equation.
Solve the Equation
Example:
Check:
9x2 = 27x
9x2 – 27x = 0
9x(x – 3) = 0
Put in standard form
GCF = 9x
9x = 0
9x/9 = 0/9
x=0
and
9x2 = 27x
9(0)2 = 27(0)
0=0
x–3=0
x-3+3 = 0+3
x=3
9x2 = 27x
9(3)2 = 27(3)
9(9) = 81
81 = 81
Solve the Equation
Example:
Check:
x2 + 11x + 34 = 4
x2 + 11x + 30 = 0
(x + 5)(x + 6) = 0
x+5=0
and
x=5–5=0–5
x = -5
x2 + 11x + 34 = 4
(-5)2 + 11(-5) + 34 = 4
25 – 55 + 34 = 4
-30 + 34 = 4
4=4
Put in standard form
Factors 30 add 11
(5)(6)
6+5
x+6=0
x+6–6=0-6
x = -6
x2 + 11x + 34 = 4
(-6)2 + 11(-6) + 34 = 4
36 – 66 + 34 = 4
-30 + 34 = 4
Solve the Equation
Example:
5z2 + 10z – 60 = -10z
Put in standard form
5z2 + 10z +10z – 60 = 0
GCF = 5
5z2 + 20z – 60 = 0
Factors of -12
add 4
5(z2 + 4z – 12) = 0
(-2)(6)
6 + -2
5(x + 6)(x – 2)
x+6=0
and x – 2 = 0
x + 6 – 6 = 0 – 6 and x – 2 + 2 = 0 + 2
x = -6
x=2
Example:
-x2 + x + 6 = 0
-1(x2 – x – 6) = 0
-1(x + 2)(x – 3) = 0
x+2=0
x+2–2=0–2
x = -2
REMEMBER:
Factor out -1 to make
squared term positive
Factors of -6
add -1
(2)(-3)
2 + -3
and
and
x–3=0
x–3+3=0+3
x=3
We can check by putting the solutions back into the
original equation.
Solve the Equation
Example:
Example:
REMEMBER:
x2 = 81
Put in standard form
2
x – 81 = 0
Difference in Two Squares
a2 – b2 = (a + b) (a – b)
(x)2 – (9)2 = 0
(x + 9) (x – 9) = 0
x + 9 = 0 and x – 9 = 0
x = -9
x=9
(x – 5)(x + 2) = 8
x2 + 2x – 5x – 10 = 8
x2 – 3x – 10 = 8
x2 – 3x – 10 – 8 = 0
x2 – 3x – 18 = 0
(x + 3)(x – 6) = 0
x + 3 = 0 and x – 6 = 0
x = -3
x=6
Multiply first,
then put in standard form
Factors of -18
add -3
(-6)(3)
-6 + 3
We can check by putting the solutions back into
the original equation.
REMEMBER
 Always put the equation in standard
form. ax2 + bx + c = 0
 The expression that is to be factored
must be equal to 0 before the zero
factor property can be used.
 Check your answer to the original
equation.
HOMEWORK 5.6
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