# Math 90 Notes Fall 2008 by 8le4LZ9l

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```									                                                                                                     Math 90

14.1 Multiplying and Simplifying Rational Expressions

A.   Rational Expressions and Replacements

P
* A rational expression is an expression that can be written in the form         ; where P
Q
and Q are polynomials.

x4
Ex.   Find the numerical value of          when x  3 .
2x  2
 4       
2   2

* Because rational expressions indicate division, we must be careful to avoid
denominators of zero. When a variable is replaced with a number that produces a
denominator equal to zero, the rational expression is not defined.

Ex.   Find all numbers for which the given rational expression is not defined.

x4
x 2  3 x  10
*Note* The value of the numerator has no bearing on
whether or not a rational expression is
defined. To determine which numbers make
the rational expression not defined, we set
the denominator equal to 0 and solve.

B.   Multiplying by 1

* We multiply rational expressions in the same way that we multiply fraction notation in
arithmetic.

Multiplying Rational Expressions: To multiply rational expressions, multiply
numerators and multiply denominators.
A C        AC
     
B D       BD

For example,
x2 x2
     
 x 2 x  2
3   x7          3 x  7
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Math 90

* Any rational expression with the same numerator and denominator is a symbol for 1.

19                   x2                     3x 2  4
 1,                   1,                         1
19                   x2                     3x 2  4

Ex.
3x  2
1 
3x  2 2x
   
 3x  2 2 x
x 1        x 1 2x                 x  1 2 x

Ex.
x2 x3
    
 x  2 x  3
x7 x3             x  7 x  3

Ex.
2  x 1
   
 2  x   1 
2  x 1         2  x   1 

C.   Simplifying Rational Expressions

Ex.   Simplify each rational expression. ( by factoring and reducing )

15    35    3
1.                
20   225   4

8x2    8 x x   x
2.                    
24 x   38 x    3

3.
x2  9

 x  3 x  3    
x3
x2  x  6        x  3 x  2        x2

5x  5
4.              
x3  x2
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Math 90

x2  8x  7
5.                
x2  4x  5

x2  4x  4
6.                
x2  2x

x4  x3
7.            
5x  5

x 2  11 x  18
8.                    
x2  x  2

x 2  10 x  25
9.                    
x2  5x

x7
10.            
x 2  49

x7
11.       
7x

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Math 90

D.   Multiplying and Simplifying

Ex.   Multiply and simplify.

x  3 2 x  10
12.                  
x  5 x2  9

7 x 2 3 y 5
13.                  
5y     14 x 2

x2  x     6
14.                   
3x      5x  5

3x  3    2x2  x  3
15.                          
5x2  5x    4x2  9

x2  6x  9 x  2
16.                     
x2  4    x3

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Math 90

14.2 Division and Reciprocals

A.   Finding Reciprocals

* Two expressions are reciprocals of each other if their product is 1. The reciprocal of a
rational expression is found by interchanging the numerator and the denominator.

2   5         2 5
1. The reciprocal of     is . (Note:    1)
5   2         5 2
2x2  3     x4            2x2  3    x4
2. The reciprocal of           is        . (Note:                   1)
x4       2x  3
2
x4      2x2  3

B.   Division

* We divide rational expressions in the same way that we divide fraction notation in
arithmetic.

Dividing Rational Expressions: To divide by a rational expression, multiply by its
reciprocal and then factor and, if possible, simplify.
             
B     D     B C       BC

Ex.   Divide and simplify.

2  3
1.         
x  x

3x3  4x3
2.          2 
40    y

4      28 x
3.              2     
x  7x
2
x  49

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6x  2   3x 2  x
4.                    
x2 1     x 1

x 1     x 1
5.          2      
x 1 x  2x 1
2

2 x 2  11 x  5   4x  2
6.                            
5 x  25        10

x2  2x  3   x 1
7.                     
x2  4      x5

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Math 90

14.6 Solving Rational Equations

A.   Rational Equations

* In Sections 14.1 and 14.2, we studied operations with rational expressions. These
expressions have no equal signs. We can multiply, divide, and/or simplify expressions,
but we cannot solve if there are no equal signs. Most often, the result of our previous
calculations is another rational expression that has not been cleared of fractions.
Equations, on the other hand, do have equals signs, and we can clear them of fractions.
A rational equation is an equation containing one or more rational expressions.

Solving Rational Equations: To solve a rational equation, the first step is to clear the
equation of fractions. To do this, multiply all terms on
both sides of the equation by the LCM of all the
denominators in the equation. Then carry out the
equation-solving process.

Finding Least Common Multiples

Ex.   Find the LCM of the denominators of each pair of rational expressions.

1      3                                     7           6
,                                             ,
8      22                                    5x        15 x 2

8222                                       5x  5  x
22  2  11                                   15 x 2  3  5  x  x

 LCM  2  2  2  11  88                   LCM  3  5  x  x  15 x 2

7x          5x2                               6m 2                   2
,                                                 ,
x2          x2                             3 m  15           m  5
2

x2                                          3 m  15 
x2                                          m  5        
2

 LCM                                        LCM 
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t  10           t5                            y5                  y4
,                                                ,
t t 6
2
t  3t  2
2
y  2y 3
2
y  3y  2
2

t2 t 6                                       y2  2y  3 
t 2  3t  2                                   y2  3y  2 

 LCM                                         LCM 

2   5   x
Ex.   Solve:                 *Note* The LCM of all denominators is 2  3  3 , or 18. We
3   6   9
multiply all terms on both sides by 18.
 2    5         x 
18      18       
 3    6         9 
2       5        x
18  18  18
3       6       9
12  15  2 x
27  2 x
27
x
2

x   x   1                               x   x   1
Ex.   Solve:                        Ex.    Solve:         
6   8 12                                4   6   8

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* If any denominator in the original equation contains a variable, you must be sure that
the proposed solution does not make any of the rational expressions, in the original
equation, undefined. That is, the solution cannot make any denominator 0 in the
original equation. Check the solution!

1    1
Ex.   Solve:                   *Note* The LCM of all denominators is x  4  x  . We multiply
x   4x
all terms on both sides by x  4  x  .
1                 1
x 4  x    x 4  x  
x                4x
4xx
4  2x
x2

1    1                                 2    1
Ex.   Solve:                          Ex.    Solve:           10
x   6x                                3x   x

6                                         1
Ex.   Solve: x       5              Ex.    Solve: x        2
x                                         x

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Math 90

x2    4                      x2      1
Ex.   Solve:                Ex.   Solve:      
x2   x2                    x 1   x 1

4     1     26                     3     1      2
Ex.   Solve:             2       Ex.   Solve:             2
x2   x2  x 4                    x5   x5  x  25

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Math 90

14.7 Applications Using Rational Equations and Proportions

A.   Solving Applied Problems

Problems Involving Work:

Ex.    Erin and Tara work as volunteers at a community recycling depot. Erin can sort a
morning‟s accumulation of recyclables in 4 hr, while Tara requires 6 hr to do the
same job. How long would it take them, working together, to sort the recyclables?

Ex.    By checking work records, a contractor finds that it takes Eduardo 6 hr to construct
a wall of a certain size. It takes Yolanda 8 hr to construct the same wall. How
long would it take if they worked together?

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Math 90

Problems Involving Motion:

* Problems that deal with distance, speed (or rate), and time are called motion
problems. Translation of these problems involves the distance formula, d  r  t .

Ex.   A zebra can run 15 mph faster than an elephant. A zebra can run 8 mi in the same
time that an elephant can run 5 mi. Find the speed of each animal.

Ex.   Nancy drives 20 mph faster than her father, Greg. In the same time that Nancy
travels 180 mi, her father travels 120 mi. Find their speeds.

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B.   Applications Involving Proportions

A    C
* An equality of ratios,       , is called a proportion. The numbers within a
B    D
proportion are said to be proportional to each other.

Ex.   A 2004 Toyota Prius is a gasoline-electric car that travels 240 mi in city driving on
4 gal of gas. Find the amount of gas required for 360 mi of city driving.

Ex.   To determine the number of fish in a lake, a park ranger catches 225 fish, tags
them, and throws them back into the lake. Later, 108 fish are caught, and 15 of
them are found to be tagged. Estimate how many fish are in the lake.

Similar Triangles:

* In similar triangles, corresponding angles have the same measure and the lengths of
corresponding sides are proportional.

Ex.   Triangles ABC and XYZ are similar triangles. Solve for z.
Y
B
5          8                    z               10

A                         C        X                             Z

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Math 90

A.   Square Roots

* When we raise a number to the second power, we have squared the number.
Sometimes we may need to find the number that was squared. We call this process
finding the square root of a number.

Square Root: The number c is a square root of a if c 2  a .

* Every positive number has two square roots. For example, the square roots of 25 are 5
and -5 because 5 2  25 and   5   25 . The positive square root is also called the
2

principal square root. The symbol             is called a radical symbol. The radical
symbol represents only the principal square root. Thus,     25  5 . To name the
negative square root of a number, we use           . The number 0 has only one square
root, 0.

Ex.   Find the square roots of 81.

The square roots are 9 and - 9.

Ex.   Find   225 .

There are two square roots of 225, 15 and - 15. We want the principal, or positive,
square root since this is what       represents. Thus, 225  15 .

Ex.   Find  64 .

The symbol     64 represents the positive square root. Then  64 represents the
negative square root. That is,    64  8 , so  64   8 .

C.   Applications of Square Roots

Ex.   After an accident, how do police determine the speed at which the car had been
traveling? The formula r  2 5 L can be used to approximate the speed r, in
miles per hour, of a car that has left a skid mark of length L, in feet. What was the
speed of a car that left skid marks of length (a) 30 ft? (b) 150 ft?
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a) We substitute 30 for L and find an approximation:

r  2 5 L  2 5  30  2 150  24.495 .

The speed of the car was about 24.5 mph.

b) We substitute 150 for L and find an approximation:

r  2 5 L  2 5  150  54.772

The speed of the car was about 54.8 mph.

* When an expression is written under a radical, we have a radical expression. The

E.   Expressions That Are Meaningful as Real Numbers

* The square of any nonzero number is always positive. There are no real numbers that
when squared yield negative numbers. Thus the following expressions do not represent
real numbers (they are meaningless as real numbers):
100 ,       49 ,     3

represent real numbers.

* In general, when replacements for x are considered to be any real numbers, it follows
that     x2  x ,

and when x  3 or x   3 ,

x2     32  3  3             x2      3         3  3
2
and

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Principle Square Root of A 2 : For any real number A ,               A 2  A . (That is, for any
real number A , the principal square root of A 2 is the
absolute value of A .)

Ex.   Simplify. Assume that expressions under the radicals represent any real number.

1.      10 2  10  10

 7         7  7
2
2.

 3x         3x
2
3.

a2b2           ab          ab
2
4.

x2  2x 1             x  1         x 1
2
5.

* Fortunately, in many cases, it can be assumed that radicands that are variable
expressions do not represent the square of a negative number. When this assumption is
made, the need for absolute-value symbols disappears.

Principal Square Root of A 2 : For any nonnegative real number A , A 2  A . (That
is, for any nonnegative real number A , the principal
square root of A 2 is A .)

Ex.   Simplify. Assume that radicands do not represent the square of a negative number.

 3x         3x
2
6.

a2b2           ab          ab
2
7.

x2  2x 1             x  1         x 1
2
8.

Radicals and Absolute Value: Henceforth, in this text we will assume that no
radicands are formed by raising negative quantities to
even powers.
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16.2 Multiplying and Simplifying with Radical Expressions

A.   Simplifying by Factoring

* To see how to multiply with radical notation, consider the following.

a)       9     4  3 2  6

b)       94         36  6

* Note that   9      4     94

The Product Rule for Radicals: For any nonnegative radicands A and B ,
A  B  A  B . (The product of square roots is
equal to the square root of the product of the

Ex.   Multiply.

1.       5      7     57 

2.       8      8     88          

2    4        2 4
3.                          
3    5        3 5

4.       2x      3x  1       2x  3x  1 

* To factor radical expressions, we can use the product rule for radicals in reverse.

Factoring Radical Expressions:           A B        A    B

* In some cases, we can simplify after factoring. When simplifying a square-root radical
expression, we first determine whether the radicand is a perfect square. Then we
determine whether it has perfect-square factors. The radicand is then factored and the
radical expression simplified using the preceding rule.

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Compare the following:

50  10  5  10             5

50         25  2     25    2 5          2

* In the second case, the radicand has the perfect-square factor 25. Thus, it is simplified.
such as 5 2 , are considered to be in simplest form.

Ex.     Simplify by factoring.

5.           18        92      9     2 

6.           48 t  16  3  t 

7.           20 t 2        45t2 

x2  6x  9          x  3       
2
8.

9.           36 x 2        36    x2 

3x 2  6 x  3       3  x2  2x 1     3  x 1 
2
10.

B.   Simplifying Square Roots of Powers

* To take the square root of an even power such as x 10 , we note that x 10   x 5  . Then
2

x 
2
x 10          5
 x 5 . We can find the answer by taking half the exponent. That is,

x 10  x 5 .

Ex.     Simplify.

11.          x6  x3

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12.      x8  x4

13.      t 22 

* If an odd power occurs, we express the power in terms of the largest even power. Then
we simplify the even power as in Examples 11-13.

Ex.   Simplify by factoring.

14.      x9        x8  x 

15.      32 x 15  16  2  x 14  x 

16.      24 x 11 

C.   Multiplying and Simplifying

* Sometimes we can simplify after multiplying. We leave the radicand in factored form
and factor further to determine perfect-square factors. Then we simplify the perfect-
square factors.

Ex.   Multiply and then simplify by factoring.

17.      2      14      2  14     28       47 2     7

18.      3      6 

19.      2      50 

20.      3x 2       9x3       3x 2  9 x3    27 x 5    9  3  x 4  x  3x 2   3x

21.      2x3        8x3 y 4 

22.      20 c d 2      35 c d 5 

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* To see how to divide with radical notation, consider the following.

25       5
a)            
16       4

25   5
b)          
16   4

25        25
* Note that         
16        16

The Quotient Rule for Radicals: For any nonnegative number A and any positive
A        A
number B,                 . (The quotient of two
B       B
square roots is equal to the square root of the quotient

Ex.   Divide and simplify.

27           27
1.                         9 3
3            3

96
2.            
6

75
3.            
3

30 a 5        30 a 5
4.                              5a 3    5 a2  a  a   5a
6a 2         6a 2

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42 x 5
5.                   
7x2

B.   Square Roots of Quotients

* To find the square root of certain quotients, we can reverse the quotient rule for
radicals. We can take the square root of a quotient by taking the square roots of the
numerator and the denominator separately.

Square Roots of Quotients: For any nonnegative number A and any positive number B,
A       A
      . (We can take the square roots of the
B       B
numerator and the denominator separately.)

Ex.   Simplify by taking the square roots of the numerator and denominator separately.

25           25       5
6.                           
9            9        3

49
7.                           
t2

* Sometimes a rational expression can be simplified to one that has a perfect-square
numerator and a perfect-square denominator.

Ex.   Simplify.

18             92          9    3
8.                                     
50             25  2       25   5

18              2
9.                                    
32              2

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48 x 3        48 x 3          16   4
10.                                       2
3x7          3x 7            x4  x

98 y
11.                                       
11
2y

C.    Rationalizing Denominators

* Sometimes in mathematics it is useful to find an equivalent expression without a radical
in the denominator. This provides a standard notation for expressing results. The procedure for
finding such an expression is called rationalizing the denominator.

Ex.   Rationalize the denominator.

2         2       2           3       6            6
12.                                             
3         3       3           3       9           3

3         3       3
13.                                                 
5         5       5

5
14.          
18

8
15.         
7

3
16.            
2

5
17.            
x

49 a 5
18.                 
12
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16.4 Addition, Subtraction, and More Multiplication

* We can add any two real numbers. The sum of 5 and        2 can be expressed as
5    2 . We cannot simplify this unless we use rational approximations such as
5  2  5  1.414  6.414 . However, when we have like radicals, a sum can be
simplified using the distributive laws and collecting like terms. Like radicals have the

1.     3 5  4 5 3 4 5 7 5

2.     8 5 3 5 

3.     5 2  18  5 2             92
5 2 3 2
2 2

4.     2 10  7 40 

5.       24     54 

6.       4x3  7     x 

7.       x3  x2       4x  4 

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1
8.        3            
3

B.   Multiplication

* Now let‟s multiply where some of the expressions may contain more than one term.

Ex.   Multiply.

9.        2      3              7       6  14

10.   2         3   54             3  

11.        3        x             3        x   

3             
2
12.               p           

2             
2
13.               5           

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16.6 Applications with Right Triangles

A.   Right Triangles

* A right triangle is a triangle with a 90 angle, as shown in the figure below. The
small square in the corner indicates the 90 angle.

Hypotenuse         c
a Leg
b
Leg

* In a right triangle, the longest side is called the hypotenuse. It is also the side opposite
the right angle. The other two sides are called legs. We generally use the letters a and
b for the lengths of the legs and c for the length of the hypotenuse. They are related as
follows.

The Pythagorean Theorem: In any right triangle, if a and b are the lengths of the legs
and c is the length of the hypotenuse, then a 2  b 2  c 2 .

Ex.   Find the length of the unknown side of each right triangle.

1.                                                    4

c             5

2.                                                    10

b
12

3.

10            15

a

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A.   Standard Form

The following are quadratic equations. They contain polynomials of second degree.

4x2  7x  5  0 ,
1
3t 2  t  9 ,
2
5 y 2  6 y

The quadratic equation 4 x 2  7 x  5  0 is said to be in standard form. Although the
quadratic equation 4 x 2  5  7 x is equivalent to the preceding equation, it is not in
standard form.

Quadratic Equation: A quadratic equation is an equation equivalent to an equation of
the type a x 2  b x  c  0 , a  0 , where a, b, and c are real-
number constants. We say that the preceding is the standard

Ex.   Write in standard form and determine a, b, and c.

1.       4x2  7x  5  0        a  ____ , b  ____ , c  ____

2.       4 y 2  5 y            a  ____ , b  ____ , c  ____

B.   Solving Quadratic Equations of the Type ax 2 + bx = 0

* Sometimes we can use factoring and the principle of zero products to solve quadratic
equations. Note: When c  0 and b  0 , we can always factor and use the principle of
zero products.

Ex.   Solve.

3.       7x2  2x  0

M. Ruvalcaba
Math 90

4.       4x  8x  0
2

C.   Solving Quadratic Equations of the Type ax 2 + bx + c = 0

* When neither b nor c is 0, we can sometimes solve by factoring.

Ex.   Solve.

5.       2 x 2  x  21  0

6.        y  3   y  2   6 y  3 

* Recall that to solve a rational equation, we multiply both sides by the LCM of all the
denominators in the equation. We may obtain a quadratic equation after a few steps.
When that happens, we know how to finish solving, but we must still remember to
check possible solutions because a replacement may result in division by 0.

3      5
7.                  2
x 1   x 1

M. Ruvalcaba
Math 90

17.2 Solving Quadratic Equations by Completing the Square

A.   Solving Quadratic Equations of the Type ax 2 = p

* For equations of the type a x 2  p , we first solve for x 2 and then apply the principle of
square roots, which states that a positive number has two square roots.

The Principle of Square Roots: * The equation x 2  d has two real solutions when
d  0 . The solutions are    d and       d .

* The equation x 2  d has no real-number solution
when d  0 .

* The equation x 2  0 has 0 as its only solution.

Ex.    Solve.

1 2
1.       x2  3                  2.       x 0
8

3.       3 x 2  7  0          4.     2x2  3  0

M. Ruvalcaba
Math 90

Solving Quadratic Equations of the Type  x + c  = d
2
B.

* In an equation of the type  x  c   d , we have the square of a binomial equal to a
2

constant. We can use the principle of square roots to solve such an equation.

Ex.   Solve.

 x 5        9                x  2       7
2                               2
5.                                 6.

 x  3        16              x  4        11
2                               2
7.                                 8.

* In Examples 5 through 8, the left sides of the equations are squares of binomials. If we
can express an equation in such a form, we can proceed as we did in those examples.

Ex. Solve.

9.       x 2  8 x  16  49       10.   x 2  6 x  9  64

M. Ruvalcaba
Math 90
C.   Completing the Square

* We have seen that a quadratic equation like  x  5   9 can be solved by using the
2

principle of square roots. We also noted that an equation like x 2  8 x  16  49 can be
solved in the same manner because the expression on the left side (perfect square
trinomial) is the square of a binomial,  x  4  . This second procedure is the basis for
2

a method of solving quadratic equations called completing the square.

* Suppose we have the following quadratic equation: x 2  10 x  4 . If we could add to
both sides of the equation a constant that would make the expression on the left a
perfect square trinomial, we could then solve the equation using the principle of square
roots.

* By adding 25 to the left side of the equation x 2  10 x , we can „force‟ a perfect
trinomial square to be present. Adding 25 to the left side of the equation would mean
that we would have to add the same amount to the right side as well. The equation
would be solved as follows:            x 2  10 x ______  4______
x 2  10 x  25  4  25
 x  5         29
2

x5              29
x  5             29

Completing the Square: To complete the square of an expression like x 2  b x , we
take half of the coefficient of x and square it. Then we add
that amount to both sides of the equation so that the left side
becomes a perfect square trinomial.

Ex.   Solve.

11.      x2  6x  8  0         12.   x2  6x  8  0

M. Ruvalcaba
Math 90

13.   x  4x  7  0
2
14.   x  12 x  23  0
2

15.   x 2  3 x  10  0   16.   x 2  3 x  10  0

17.   2 x 2  3x  1       18.   2 x 2  3x  3  0

M. Ruvalcaba
Math 90

A.   Solving Using the Quadratic Formula

The Quadratic Formula: The solutions of a x 2  b x  c  0 are given by

b    b 2  4 ac
x
2a

Ex.   Solve using the quadratic formula.

1.    5 x 2  8 x  3         2.    x 2  3 x  10  0

3.    x2  4x  7              4.    x 2  x  1

M. Ruvalcaba
Math 90

17.5 Applications and Problem Solving

A.   Using Quadratic Equations to Solve Applied Problems

Ex.   Solve.

1.       The area of a rectangular red raspberry patch is 76 ft 2 . The length is 7 ft
longer than three times the width. Find the dimensions of the raspberry
patch.

2.       The length of a rectangular area rug is 3 ft greater than the width. The area
is 70 ft 2 . Find the length and the width.

3.       A square is a carpenter‟s tool in the shape of a right triangle. One side, or
leg, of a square is 8 in. longer than the other. The length of the hypotenuse
is 8 13 in. Find the lengths of the legs of the square.

M. Ruvalcaba
Math 90
4.   The current in a stream moves at a speed of 2 km / h . A boat travels 24 km
upstream and 24 km downstream in a total time of 5 hr . What is the speed
of the boat in still water?

5.   The speed of a boat in still water is 10 km / h . The boat travels 12 km
upstream and 28 km downstream in a total time of 4 hr . What is the speed
of the stream?

M. Ruvalcaba
Math 90

A.       Graphing Quadratic Equations of the Type y = ax 2 + bx + c
&
B.       Finding the x-Intercepts of a Quadratic Equation

Ex. Graph: y  x 2                                          Ex. Graph: y   x 2
y
6

x   y                                         5                                         x           y
4
3

2
1

x
-6 -5   -4 -3 -2   -1    0   1   2   3   4   5   6
-1

-2

-3
-4

-5
-6

First find these 4 items of information…
1.   Vertex
2.   y-intercept
3.   Whether the parabola opens up or down
4.   x-intercept(s)

b
1. To find the vertex, use the vertex formula ( x        ) to find the x-coordinate. Then, substitute the
2a
value of the x-coordinate back into the original quadratic equation to find the y-coordinate.

2. The y-intercept(where the parabola intersects the y-axis) is simply the “c”, i.e. the constant.

3. If the “a” is positive, the parabola opens up and if the “a” is negative, the parabola opens down.

4. To find the x-intercept(s), replace the “y” with a 0 and solve the remaining equation for “x”.

Ex.      Graph: y  x 2  2 x  3                                                                                           y
6

5
4
3

2

1

x
-6 -5   -4 -3   -2   -1    0   1   2   3   4   5   6
-1
-2
-3

-4
-5

-6

M. Ruvalcaba
Math 90
Ex.   Graph the quadratic equation. Label the ordered pairs for the vertex and the y-intercept.
y
6
1.     y  x 1
2
5
4
3

2
1

x
-6 -5   -4 -3 -2     -1    0   1   2   3   4   5     6
-1
-2

-3
-4

-5
-6

y
2.     y  x2  2x                                                                  6
5

4
3

2
1

x
-6 -5   -4 -3   -2   -1    0   1   2   3   4   5    6
-1
-2

-3
-4

-5

-6

3.     y  x2  2x 1                                                                6
y
5

4
3

2
1

x
-6 -5   -4 -3   -2   -1    0   1   2   3   4   5    6
-1
-2

-3
-4

-5

-6

4.     y  x2  2x  3                                                              6
y
5
4
3

2

1

x
-6 -5   -4 -3   -2   -1    0   1   2   3   4   5    6
-1
-2
-3

-4
-5

-6

M. Ruvalcaba

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