Math 0094

Worksheet

Test 4.

Solutions

 

Solve the following equations:

 

Factor and set each factor = 0.

(x + 9)(x + 3) = 0

x + 9 = 0 or x + 3 = 0

x = -9 or x = -3

 

(3x –4) (x + 6)=0

3x – 4 = 0 or x + 6 = 0

3x = 4

x = 4/3 or x = -6

 

Simplify each side, then set = 0

 

since there is no middle term, set the square = to 27 and then take the square root of each side. 

 

  1. A rectangular porch with an area of 108 square feet is added to a house.  The width of the porch is 6 feet less than twice its depth.  What are the dimensions of the porch?

Depth = x

Width = 2x – 6

Area  = x (2x – 6)

x can’t be negative, so x = 9 feet.  Therefore the dimensions of the porch are

9 ft. x 12 ft.

 

  1. The rectangular garden is to be increased so that its new area is 162 square feet more than the original garden.  This is done by increasing both the length and width 6 feet.  The original length is 3 feet more than its original width.  What are the new dimensions of the garden?

 

Original            length = x + 3

                        width = x

                        Area = x(x+3)

New                 length = x + 9

                        width = x + 6

                        Area = (x + 9)(x + 6)

 

               X+3              6

 


            x

 

            6

 

old area + 162 = new area

The garden is now 15 ft. by 18 ft.

 

  1. A ladder is leaned up against the side of a house.  If the ladder reaches 12 feet up the house and the length of the ladder is 3 feet more than twice the distance the base of the ladder is from the house, how long is the ladder?

 

 

 


Height = 12 ft.

12
 
                                    Base = x

                                    Ladder = 2x + 3

                        2x +3

 

                    x

x = -9 or x = 5.  Since x is a distance it cannot be negative and x = 5 feet.  The ladder is 13 feet long.

 

  1. A ball is thrown into the air with an initial velocity of 38 ft per second from a height of five feet.  Its height after t seconds can be found by . 
    1. How high will the ball be in 2 seconds?  Plug in 2 for t.

H = -16(2)^2 + 38 (2) + 5

H = -16(4) + 76 + 5

H = -64 + 81

H = 17 feet in the air after 2 seconds.

 

    1. When will the ball hit the ground?  The height of the ball will be zero when it hits the ground, so plug in 0 for h.

0 = -16t^2 +38t + 5

0 = 16t^2 –38t – 5 by multiplying both sides by –1.

Factor

0 = (8t  + 1)(2t - 5)  t can’t be negative, so

2t – 5 = 0

2t = 5

t = 2.5 seconds until the ball hits the ground.

 

  1. Solve:  Graph the solution on a number line and dexcribe it using interval notation.

Factor and set each factor = 0 to find the critical points.

(x + 9)(x + 3 ) = 0

critical points are x = -9 and x = -3

Factors:        (-)(-)                                (+)(-)               (+)(+)

 


            Results: (+)            -9                (-)        -3         (+)

We want the negative regions which is only the middle area:

So the solution set is:

 

                                                           

                                                            -9                       -3

in interval notation: [-9, -3]

 

2x(x^2 – 12x +27)

2x(x - 3)(x – 9)                       

critical values x = 0, x = 3, x = 9

 

            (-)(-)(-)                           (+)(-)(-)                    (+)(+)(-)           (+)(+)(+)

 


            (-)                    0     (+)                        3         (-)               9      (+)

We want the positive regions:

                                    (                          )                       (

                                    0                       3                            9     

in interval notation: .

 

  1. Evaluate the following to the nearest tenth.
    1. = 8
    2. = 12/7
    3. = 6.1
    4. = not a real number
    5. -= -9
  2. Square the following:
    1. squared is .
  3. Simplify the following.  Do not evaluate.

2*192

2 * 2 * 96

2*2*2*48

2*2*2*2*24

2*2*2*2*2*12

2*2*2*2*2*2*6

2*2*2*2*2*2*2*3

There are 3 sets or 2’s and one 3 and one 2 left over.  Therefore it is equal to:

 

Since 72 = 36 * 2, a 6 comes out and the 2 stays in.  divide the exponents by 2 to determine how many of the variables come out. If there is a remainder, that many of the variable stays in the radical.

                       

 

256 = 2*2*2*2*2*2*2*2, so there are two sets of three 2’s and two 2’s left over.  Divide the exponents of the variable by 3 this time since it is a cube root.

 

  1. Add/Subtract the following:

Remember that you can only add if the exact same value is under the radical.

 

 

 

 

 

  1. Rationalize the following:

 

    1.                        9 is 3*3, so we need one more 3, we also need another x and 2 more y’s.

 

    1.                                    We need two more 2’s, 2 more x’s, and one more y.

 

Multiply or Divide the following:

 

 

  1. since both signs are opposites, they reduce to –1.

 

  1.