Name                          .

 

Math 152B Practice Midterm II

Please do all of the following problems.  All work and all answers must be done on your own paper.  Credit earned will be based on the steps that you show that lead to the final solution.  Good Luck!

 

 

Problem 1   Find the root: 

         

   

Solution

        -2, since (-2)3 = -8

 

 

Problem 2   Simplify the expression.  Leave your answer in exponential form with only positive exponents.

       

 Solution

       

 

Problem 3  Multiply and Simplify

 

Solution

       

 

 

 

Problem 4   Simplify the expression.  Assume all variables represent positive real number.

         

   

 Solution

       

 

 

Problem 5    Multiply and Simplify

       

   

   Solution

               

 

Problem 6  Rationalize the denominator

             5
                       
          4 -

    

 Solution

                     5    (4 + )
                                                
                 ( 4 - )(4 + )

 

                      20 + 5
             =                          
                        16  -  6

 

                      20 + 5
             =                          
                            10

 

Problem 7   Simplify

       

   

 Solution

       

 

 

Problem 8 

Write down the quadratic formula

Solution

The solutions to the equation

        ax2 + bx + c  =  0

are

        x  =    

 

 

Problem 9

Use the discriminant to determine the nature of the roots of the following quadratics.

A.     2x2 + 3x + 4  =  0

 

Solution

The discriminant is

        32 - 4(2)(4) = 9 - 32 < 0

Since the discriminant is negative, we can conclude that there are no real roots, only two complex conjugate roots.

   

 

B.    3x2 + x - 4  =  0 

 

Solution

The discriminant is

        12 - 4(3)(-4) = 1 + 48 > 0

Since the discriminant is positive, we can conclude that there are two distinct real roots.

 

 

Problem 10  

Solve the equation by using the factoring or root method.

A) 10x2  =  13x + 3 

           

Solution

First bring all terms to the left hand side

        10x2 - 13x - 3  =  0

There are two ways of completing this problem.  The first is to use the quadratic formula

        a  =  10    b  =  -13    c  =  -3

       

Hence

                13 + 17
        x =                   
                   20    

or

                13 - 17
        x =                   
                   20    

We get  x  =  3/2    or   x  =  -1/5

 

Alternatively, we can factor using the AC method:

        AC  =  -30

The two numbers that multiply to -30 and add to -13 are

        (-15,2)

We now rewrite

        10x2 - 13x - 3  =  10x2 - 15x + 2x - 3

        =  5x(2x - 3) + 1(2x - 3)        Factoring by grouping

        =  (5x + 1)(2x - 3)  =  0

Hence

        5x + 1  =  0     or       2x - 3 = 0

        x  =  -1/5    or    x  =  3/2

 

B)   2x2 - 15 = 11

 

  Solution

        To solve this, there is no "x" term, so we can use the square root property:

        2x2  =  26        Subtracting 15 from both sides

        x2  =  13        Dividing both sides by two

        x  =  sqrt(13)       Taking the square root of both sides

   

 

 

Problem 11 

Solve the following by completing the square

   5x2 - 20x - 21  =  0

Solution

        5(x2 - 4x)  - 21  =  0        Pulling out the 5 from the first two terms

Now calculate the magic number (b/2)2

        (-4/2)2  =  (-2)2  =  4

        5(x2 - 4x + 4 - 4)  - 21  =  0        Adding and subtracting the magic number 4

        5[(x - 2)2 - 4] - 21  =  0        Factoring the first three terms

        5(x - 2)2 - 20 - 21  =  0        Distributing the 5 through

        5(x - 2)2 - 42  =  0        Adding the numbers

        5(x - 2)2  =  42        Adding the numbers

        (x - 2)2  =  42/5        Dividing both sides by 5

        x - 2  =  sqrt(42/5)        Taking the square root of both sides

        x  =  2   sqrt(42/5)     Adding 2 to both sides

        x  =  2 sqrt(210) / 5        Rationalizing the denominator

 

Problem 12

Solve the following by any method

A)   6x2 + 5x - 4  =  0 

   

Solution

We use the quadratic formula with

        a  =  6,    b  =  5,    c  =  -4

       

So that 

        x   =   1/2  +  11/12         or        x   =   1/2  -  11/12   

Notice that we could have instead used the AC method to factor.

 

B)    2x3 - 18x  =  0 

   

  Solution

We factor:

        2x(x2 - 9)  =  0        Pulling out the greatest common factor

        2x(x - 3)(x + 3)  =  0        Difference of squares

        x  =  0,       x  =  3,   or        x  =  -3        Zero product formula    

 

Problem 13 

The pressure p in pounds per square foot of a wind is directly proportional to the square of the velocity v of he wind.  If a 10-mi/hr wind produces a pressure of 0.3 lb/ft2, what pressure will a 100-mi/hr wind produce?

   

Solution

        This is a variation problem.  The first sentence implies

        p  =  kv2

        The second sentence tells us

        0.3  =  k(10)2        When v  =  10  p  =  0.3

        0.3  =  100k

        k  =  0.3/100  =  0.003        Dividing both sides by 100

        p  =  0.003v2        Substituting k  =  0.003 back into the original equation

        Now we want to know what p is when v  =  100

        p  =  0.003(100)2        Substituting 100 in for v

        p  =  0.003(10,000)  =  30

        A 100 mile per hour wind will produce a pressure of 30lb/ft2 

 

 

Problem 14

Factor

        4x2 -12x + 9

Solution

Notice that this is the square of the difference with

        a  =  2x        and          b  =  3

We get

        (2x - 3)2

 

 

Problem 15

Derek bicycled 36 miles to get to Echo Summit and back and Nick bicycled 60 miles to get to Carson Pass and back.  Nick rode 3 miles per hour faster than Derek, and his trip took an hour longer than Derek's.  What is the fastest speed that Derek could have been traveling?  (You must set up the equations, that is, no guessing).

Solution

We construct the following Distance-Rate-Time table

  Distance Rate Time
Derek 36 r t
Nick 60 r + 3 t + 1

This gives us the two equations

        36  =  rt        60  =  (r + 3)(t + 1)

The first equation gives us

        t  =  36/r

so that

        60  =  (r + 3)(36/r + 1)

Multiplying both sides by r gives

        60r  =  (r + 3)(36 + r)  =  r2 + 39r + 108

or

        r2 - 21r  + 108  =  0

        (r - 9)(r - 12)  =  0

So that

        r  =  9     or      r  =  12

The fastest that Derek could have been traveling is 12 miles per hour.

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