Download MTH55_Lec-48_sec_8-1_Complete_Square

Chabot Mathematics
§8.1 Complete
The Square
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Chabot College Mathematics
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[email protected] • MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt
Review § 7.7
MTH 55
 Any QUESTIONS About
• §7.7 → Complex Numbers
 Any QUESTIONS About HomeWork
• §7.7 → HW-36
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The Square Root Property
 Let’s consider x2 = 25.
 We know that the number 25 has
two real-number square roots, 5
and −5, which are the solutions to
this equation.
 Thus we see that square roots can
provide quick solutions for
equations of the type x2 = k.
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SQUARE ROOT PROPERTY
 For any nonzero real number d, and
any algebraic expression u, then the
Equation u2 = d has exactly two
solutions:
If u  d
2
then u  d
or u   d
 Alternatively in a ShortHand Notation:
If u  d
2
Chabot College Mathematics
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then u   d
Bruce Mayer, PE
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Example  Use SqRt Property
 Solve 5x2 = 15. Give exact solutions and
approximations to three decimal places.
2
5
x
 15
 SOLUTION
x2  3
Isolating x2
Using the Property of
x  3 or x   3. square roots
Or x   3
ShortHand Notation
 The solutions are  3, which round
to 1.732 and −1.732.
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Bruce Mayer, PE
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Example  Use SqRt Property
 Solve x2 = 108
2
x
 108
 SOLN
x   108
Use the square root principle.
x   36 3
Simplify by factoring out a perfect square.
x  6 3
Note: Remember the ± means that
the two solutions are 6 3 and 6 3 .
Check
Check

6 3: 6 3

2
 108
36  3  108
Chabot College Mathematics
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

6 3 : 6 3

2
 108
36  3  108

Bruce Mayer, PE
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Example  Use SqRt Property
 Solve x2 +14 = 32
2
x
 14  32
 SOLN
x  18
Subtract 14 from both sides to isolate x2
x   18
Use the square root property
2
x   9 2
Simplify by factoring out a perfect square
x  3 2
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Example  Use SqRt Property
 Solve (x + 3)2 = 7
 SOLN ( x  3)2  7
x  3  7 or x  3   7
Using the Property
of square roots
x  3  7 or x  3  7. Solving for x
Or x  3  7
ShortHand Notation
 The solutions are 3  7. The check
is left for us to do Later
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Bruce Mayer, PE
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Example  Use SqRt Property
 Solve 16x2 + 9 = 0
 SOLN 16 x2  9  0
2
x  9/16
x   9 16 or x    9 16
3
3
x  i or x   i Recall that 1  i.
4
4
3
 The solutions are  i The check
4
is left for Later
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Bruce Mayer, PE
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Solving Quadratic Equations
 To solve equations in the form ax2 = b, first
isolate x2 by dividing both sides of the
equation by a.
 Solve an equation in the form ax2 + b = c
by using both the addition and multiplication
principles of equality to isolate x2 before
using the square root principle
 In an equation in the form (ax + b)2 = c, notice
the expression ax + b is squared. Use the
square root principle to eliminate the square.
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Bruce Mayer, PE
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Example  Use SqRt Property
 Solve (5x − 3)2 = 4
2
5
x

3
 4
 SOLN 
5x  3   4
Use the square root property
5x  3  2
2  3
x
5
23
2  3
or x 
x
5
5
1
or
x
x 1
5
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Add 3 to both sides and divide
each side by 5, to isolate x.
Bruce Mayer, PE
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Completing the Square
 Not all quadratic equations can be
solved as in the previous examples.
 By using a method called
completing the square, we can
use the principle of square roots to
solve any quadratic equation
 To Complete-the-Sq we Add ZERO
to an expression or equation
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Example  Complete the Sq
 Solve x2 + 10x + 4 = 0
 SOLN: x2  10 x  4  0
x2  10 x
x2
 4
+ 10x + 25 = –4 + 25
( x  5)2  21
x  5  21 or x  5   21
Adding 25 to
both sides.
Factoring
Using the property
of square roots
 The solutions are 5  21. The check
is left for Later
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Solving Quadratic Equations
by Completing the Square
 Write the equation in the form 1·x2 + bx = c.
 Complete the square by adding (b/2)2 to both
sides.
• (b/2)2 is called the “Quadratic Supplement”
 Write the completed square in factored form.
 Use the square root property to eliminate the
square.
 Isolate the variable.
 Simplify as needed.
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Bruce Mayer, PE
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Example  Complete the Sq
 Solve by Completing the Square:
2x2 − 10x = 9
2
2
x
 10 x  9
 SOLN:
2 x 2  10 x 9

2
2
9
2
x  5x 
2
25 9 25
2
x  5x 
 
4 2 4
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Divide both sides by 2.
Simplify.
25
Add
to both sides to complete
4
the square.
Bruce Mayer, PE
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Example  Complete the Sq
 Solve 2x2 − 10x = 9
2
 SOLN:  x  5   18  25  43

2
4
4
5
43
x 
2
4
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4
Use the square root principle.
5
43
x 
2
2
5  43
x
2
Factor.
5
Add 2 to both sides and simplify
the square root.
Combine the fractions.
Bruce Mayer, PE
[email protected] • MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt
Example  Complete the Sq
 Solve by Completing the Square:
3x2 + 7x +1 = 0
 SOLUTION: The coefficient of the x2
term must be 1. When it is not, multiply
or divide on both sides to find an
equivalent eqn with an x2 coefficient of 1.
3x  7 x  1  0
2
1
3
1
3
x

7
x

1


 3 0
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2
Divide Eqn by 3
Bruce Mayer, PE
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Example  Complete the Sq
 Solve: 3x2 + 7x +1 = 0
7
1
2
 SOLN:
x  x 0
3
3
7
x  x
3
2
1

3
7
49
1 49
x  x
 
3
36
3 36
2
2
7
12 49

x   
6
36 36

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Bruce Mayer, PE
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Example  Complete the Sq
 Solve: 3x2 + 7x +1 = 0
2
 SOLN:  x  7   37



6
36
2
7
37

x  
6
36

7
37
x 
6
6
7
37
or x   
6
6
Square Root
Property
7
37
x 
6
6
7
37
or x   
6
6
Isolate
x
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Taking the Square
Root of Both Sides
Bruce Mayer, PE
[email protected] • MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt
Example  Taipei 101 Tower
 The Taipei 101 tower in Taiwan is 1670
feet tall. How long would it take an
object to fall to the ground from the top?
 Familiarize: A formula for
Gravity-Driven FreeFall with negligible
air-drag is s = 16t2
• where
– s is the FreeFall Distance in feet
– t is the FreeFall Time in seconds
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Bruce Mayer, PE
[email protected] • MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt
Example  Taipei 101 Tower
 Translate: We know the distance is
1670 feet and that we need to solve for t
Sub 1670 for s → 1670 = 16t2
 CarryOut: 1670 = 16t2
1670 2
t
16
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1670
t
16
or
1670

t
16
10.2  t
or
10.2  t
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Example  Taipei 101 Tower
 Check: The number −10.2 cannot be a
solution because time cannot be
negative.
• Check t = 10.2 in formula:
s = 16(10.2)2 = 16(104.04) = 1664.64
– This result is very close to the 1670 value.
 State. It takes about 10.2 seconds for
an object to fall to the ground from the
top of the Taipei 101 tower.
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Bruce Mayer, PE
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Compound Interest
 After one year, an amount of money P,
invested at 4% per year, is worth 104% of P,
or P(1.04). If that amount continues to earn
4% interest per year, after the second year
the investment will be worth 104% of
P(1.04), or P(1.04)2. This is called
compounding interest since after the first
period, interest is earned on both the initial
investment and the interest from the first
period. Generalizing, we have the following.
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Bruce Mayer, PE
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Compound Interest Formula
 If an amount of money P is
invested at interest rate r,
compounded annually, then in t
years, it will grow to the amount A
as given by the Formula
A  P(1  r ) .
t
• Note that r is expressed as a DECIMAL
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Bruce Mayer, PE
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Example  Compound Interest
 Tariq invested $5800 at an interest rate
of r, compounded annually. In two
years, it grew to $6765.
What was the interest rate?
 Familiarize: This is a compound
interest calculation and we are already
familiar with the compound-interest
formula.
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Bruce Mayer, PE
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Example  Compound Interest
 Translate The translation consists of
substituting into the Interest formula
t
A  P(1  r )
6765 = 5800(1 + r)2
 CarryOut: Solve for r
6765/5800 = (1 + r)2
 6765/5800  1  r
1  6765/5800  r
r  .08 or r  2.08
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Bruce Mayer, PE
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Example  Compound Interest
 Check: Since the interest rate can NOT
negative, we need only to check 0.08 or
8%.
 If $5800 were invested at 8%
compounded annually, then in 2 yrs it
would grow to 5800·(1.08)2, or $6765.
• The number 8% checks.
 State: Tariq’s interest rate was 8%.
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Bruce Mayer, PE
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Solving Formulas
 Recall that to solve a formula for a
certain letter-variable, we use the
principles for solving equations to
isolate that letter-variable alone on
one side of the Equals-Sign
• The Bernoulli
Equation for an
InCompressible
Fluid:
Chabot College Mathematics
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2
V
P
z
C
2g
g
Bruce Mayer, PE
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Example  Solve

1 2
 SOLN D  n  5n
2
2D  n2  5n
2


Multiplying both sides by 2
5
5
n  5n     2 D   
2
2
2
Complete the Square
2
5
25 8D  25


 n    2D 
2
4
4

2
5  8D  25
n
2
Chabot College Mathematics
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
1 2
D  n  5n for n.
2
Express LHS as
Perfect Square
Solve Using Square
Root Principle
Bruce Mayer, PE
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Solve a Formula for a Letter – Say, b
1. Clear fractions and use the principle of powers,
as needed. Perform these steps until radicals
containing b are gone and b is not in any
denominator.
2. Combine all like terms.
3. If the only power of b is b1, the equation can be
solved without using exponent rules.
4. If b2 appears but b does not, solve for b2 and
use the principle of square roots to solve for b.
5. If there are terms containing both b and b2,
put the equation in standard form and
Complete the Square.
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Bruce Mayer, PE
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WhiteBoard Work
 Problems From §8.1 Exercise Set
• 22, 44, 56, 78, 88
 Solve ax2 + bx + c = 0 by
completing the square:
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Bruce Mayer, PE
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All Done for Today
Taipei 101
Tower
Taipei, R.o.C.
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Bruce Mayer, PE
[email protected] • MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
–
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt
Graph y = |x|
6
 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
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5
y = |x |
6
5
4
3
2
1
0
1
2
3
4
5
6
y
4
3
2
1
x
0
-6
-5
-4
-3
-2
-1
0
1
2
3
-1
-2
-3
-4
-5
file =XY_Plot_0211.xls
-6
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4
5
6
5
5
y
4
4
3
3
2
2
1
1
0
-10
-8
-6
-4
-2
-2
-1
0
2
4
6
-1
0
-3
x
0
1
2
3
4
5
-2
-1
-3
-2
M55_§JBerland_Graphs_0806.xls
-3
Chabot College Mathematics
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-4
M55_§JBerland_Graphs_0806.xls
-5
Bruce Mayer, PE
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