Math 90 Notes Fall 2008 by 8le4LZ9l

VIEWS: 62 PAGES: 36

									                                                                                                     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
                                                                                                 M. Ruvalcaba
                                                                                        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
                                                                                    M. Ruvalcaba
                              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




                          M. Ruvalcaba
                                                 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




                                             M. Ruvalcaba
                                                                                            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.
                                     A     C     A D       AD
                                                     
                                     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




                                                                                        M. Ruvalcaba
                                       Math 90



     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




                                   M. Ruvalcaba
                                                                                              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 
                                                                                          M. Ruvalcaba
                                                                                             Math 90


        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




                                                                                       M. Ruvalcaba
                                                                                           Math 90


* 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




                                                                                       M. Ruvalcaba
                                                                    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




                                                                M. Ruvalcaba
                                                                                            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?




                                                                                        M. Ruvalcaba
                                                                                       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.




                                                                                 M. Ruvalcaba
                                                                                            Math 90


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




                                                                                        M. Ruvalcaba
                                                                                            Math 90

                     16.1 Introduction to Radical Expressions

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?
                                                                                        M. Ruvalcaba
                                                                                           Math 90


             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.


D.   Radicands and Radical Expressions

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


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

     Excluding Negative Radicands: Radical expressions with negative radicands do not
                                   represent real numbers.


F.   Perfect-Square Radicands

     * 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


                                                                                       M. Ruvalcaba
                                                                                                Math 90



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.
                                                                                            M. Ruvalcaba
                                                                                             Math 90

           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
                                    radicands.)

     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.

                                                                                         M. Ruvalcaba
                                                                                                       Math 90


     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.
       Square-root radical expressions in which the radicand has no perfect-square factors,
       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

                                                                                                   M. Ruvalcaba
                                                                                                       Math 90



           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 

                                                                                                   M. Ruvalcaba
                                                                                         Math 90

                   16.3 Quotients Involving Radical Expressions

A.   Dividing Radical Expressions

     * 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
                                     of the radicands.)

     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


                                                                                     M. Ruvalcaba
                                                                                            Math 90



                       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



                                                                                        M. Ruvalcaba
                                                                                             Math 90



                      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
                                                                                         M. Ruvalcaba
                                                                                            Math 90

                16.4 Addition, Subtraction, and More Multiplication

A.   Addition and Subtraction

     * 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
      same radicands.


     Ex.   Add or subtract. Simplify, if possible, by collecting like radicals.

           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 




                                                                                        M. Ruvalcaba
                                                                                       Math 90



                                 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           




                                                                                   M. Ruvalcaba
                                                                                               Math 90

                        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


                                                                                           M. Ruvalcaba
                                                                                           Math 90

                       17.1 Introduction to Quadratic Equations

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
                         form of a quadratic equation.

     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

                            17.3 The Quadratic Formula

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

                                17.6 Graphs of Quadratic Equations

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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