Transcript 5.2: Solving Quadratic Equations by Factoring

1.3 Solving Quadratic
Equations by Factoring
(p. 18)
How can factoring be used to solve quadratic
equation when a=1?
Vocabulary Reminder
• Monomial
An expression that is either a
number, a variable, or the
product of a number and one
or more variables. (4, x, 5y)
• Binomial
The sum of two monomials (x+4)
• Trinomial
To solve a quadratic eqn.
by factoring, you must
remember your factoring
patterns!
Factor the expression.
a.
x2 – 9x + 20
SOLUTION
a. You want x2 – 9x + 20 = (x + m) (x + n) where
mn = 20 and m + n = – 9.
ANSWER
Notice that m = – 4 and n = – 5.
So, x2 – 9x + 20 = (x – 4)(x – 5).
Factor b. x2 + 3x – 12
b. You want x2 + 3x – 12 = (x + m) (x + n) where
mn = – 12 and m + n = 3.
ANSWER
Notice that there are no factors m and n such
that m + n = 3. So, x2 + 3x – 12 cannot be
factored.
Guided Practice
Factor the expression. If the expression cannot
be factored, say so.
1. x2 – 3x – 18
SOLUTION
You want x2 – 3x – 18 = (x + m) (x + n) where mn
= – 18 and m + n = – 3.
Factor of
– 18 : m, n
Sum of
factors:
m+n
1, – 18 –1, 18
– 17
17
– 3, 6
– 2, 9
2, – 9
– 6, 3
3
7
–7
–3
ANSWER
m = – 6 and n = 3 so x2 – 3x – 18 = (x – 6) (x + 3)
Guided Practice
2.
n2 – 3n + 9
SOLUTION
You want n2 – 3n + 9 = (x + m) (x + n) where mn
= 9 and
m + n = – 3.
Factor of
9 : m, n
Sum of
factors:
m+n
1, 9
10
– 1, – 9
3, 3
– 3, – 3
–10
6
–6
ANSWER
Notice that there are no factors m and n such that
m + n = – 3 . So, n2 – 3x + 9 cannot be factored.
Factor with Special Patterns
Factor the expression.
a. x2 – 49 = x2 – 72
= (x + 7) (x – 7) Difference of two squares
b. d 2 + 12d + 36 = d 2 + 2(d)(6) + 62
= (d + 6)2
Perfect square trinomial
c.
z2 – 26z + 169 = z2 – 2(z) (13) + 132
= (z – 13)2
Perfect square trinomial
Guided Practice
Factor the expression.
4. x2 – 9 = x2 – 32
Difference of two squares
= (x – 3) (x + 3)
5.
q2 – 100 = q2 – 102 Difference of two squares
= (q – 10) (q + 10)
6.
y2 + 16y + 64 = y2 + 2(y) 8 + 82
= (y + 8)2
Perfect square trinomial
Zero Product Property
• Let A and B be real numbers or algebraic
expressions. If AB=0, then A=0 or B=0.
• This means that If the product of 2 factors
is zero, then at least one of the 2 factors
had to be zero itself!
• The answers, the solutions of a quadratic
equation are called the roots or zeros of
the equation.
Use a Quadratic Equation as a Model
Nature Preserve
A town has a nature preserve
with a rectangular field that
measures 600 meters by 400
meters. The town wants to
double the area of the field by
adding land as shown. Find the
new dimensions of the field.
SOLUTION
480,000 = 240,000 + 1000x + x2 Multiply using FOIL.
Write in standard form.
0 = x2 + 1000x – 240,000
Factor.
0 = (x – 200) (x + 1200)
Zero product property
x – 200 = 0 or x + 1200 = 0
x = 200 or x = – 1200
Solve for x.
ANSWER
Reject the negative value, – 1200. The field’s
length and width should each be increased by 200
meters. The new dimensions are 800 meters by
600 meters.
What If ? In Example 4, suppose the field
initially measures 1000 meters by 300 meters.
Find the new dimensions of the field.
SOLUTION
New Area =
2(1000)(300) =
New
Length
(meters)
(1000 + x)
New width
(meters)
(300 + x)
600000 = 300000 + 1000x + 300x + x2 Multiply using FOIL.
0 = x2 + 1300x – 300000
Write in standard form.
Factor.
0 = (x – 200) (x + 1500)
x – 200 = 0 or x + 1500 = 0 Zero product property
Solve for x.
x = 200 or x = – 1500
ANSWER
Reject the negative value, – 1200. The field’s
length and width should each be increased by
200 meters. The new dimensions are 1200
meters by 500 meters.
Finding the Zeros of an Equation
• The Zeros of an equation are the x-intercepts !
•
•
•
•
First, change y to a zero.
Rewrite the function in intercept form.
Now, solve for x.
The solutions will be the zeros of the equation.
Example: Find the Zeros of
y=x2-x-6
y=x2-x-6
0=x2-x-6
0=(x-3)(x+2)
x-3=0 OR x+2=0
+3 +3
-2 -2
x=3 OR x=-2
Change y to 0
Factor the right side
Set factors =0
Solve each equation
Check your solutions!
If you were to graph the eqn., the graph would
cross the x-axis at (-2,0) and (3,0).
Guided Practice
Find the zeros of the function by rewriting the
function in intercept form.
y = x2 + 5x – 14
SOLUTION
y = x2 + 5x – 14
= (x + 7) (x – 2)
Write original function.
Factor.
The zeros of the function is – 7 and 2
Check Graph y = x2 + 5x – 14. The graph passes
through ( – 7, 0) and (2, 0).
Assignment
p. 21, 3-54 every 3rd problem
(3,6,9,12,…)