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EDUCATIONAL GAMES SOLVE A QUADRATIC EQUATION BY COMPLETING THE SQUARE
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College Algebra
Answer/Discussion to Practice Test 


Problems 1a - 1c:  Solve the given equation and determine whether the equation is an identity, a conditional equation or an inconsistent equation.

 
1a. 
 
 
Answer:

Since the variable dropped out AND it is a true statement, the answer is all real numbers, which means this equation is an identity.


 
1b. 
 
 
Answer:

Since we came up with the solution x = 29/36, this would be an example of a conditional equation.


 
1c. 
 
 
Answer:

Since the variable dropped out AND it is a false statement, the answer is no solution which means this is an inconsistent equation.


 
 
Problems 2a - 2b:  Solve the given equation.

 
2a. 
 
 
Answer:

Note that -2 does not cause any denominators to be zero.  So it is not an extraneous solution. 

-2 is the solution to our equation.


 
2b. 
 
 
Answer:

Note that 3 does cause two of the denominators to be zero. 

So 3 is an extraneous solution.  That means there is no solution. 

The answer is NO solution.


 
 
Problems 3a - 3c:  Solve the given word problem.

 
3a. In last weeks football game, Ralph scored 6 less than twice what Tommy scored.  The sum of their scores is 30.  How many points did Ralph and Tommy score individually?
 
 
Answer:
Assign variable:
 x = Tommys score
2x - 6 = Ralphs score

Since their sum is 30, we get the following equation:

Solving the equation we get:

 

Answer:
Tommy scored 12 points.
Ralph scored 2(12) - 6 = 18 points.


 
3b.  The ages of three sisters are three even consecutive integers.  If the sum of the 1st, four times the 2nd, and twice the 3rd is 86, what are the three ages?
 
 
Answer:
Assign variable:
x = 1st even consecutive integer
x + 2 = 2nd even consecutive integer
x + 4 = 3rd even consecutive integer

Since the sum of the 1st, four times the 2nd, and twice the 3rd is 86, we get the following equation:

Solving the equation we get:

 

Answer:
The three ages are 10, 12, and 14.


 
3c.  You are buying a computer at a markdown price of $960.  If the markdown price was 20% of the original price, how much was the computer originally?
 
 
Answer:
Assign variable:
x = original price

Since the markdown price is $960, we get the following equation:

Solving the equation we get:

 

Answer:
The original price of the computer is $1200.


 
 
Problem 4a:  Solve for the given variable.

 
4a.  for y.
 
 
Answer:

 
 
Problems 5a - 5b: Solve the given equation by factoring.

 
5a. 
 
 
Answer:

There are two solutions to this equation: x = -2 and x = 7.


 
5b. 
 
 
Answer:

There are two solutions to this equation: x = -1/2 and x = -2/3.


 
 
Problems 6a - 6b:  Solve the given equation by completing the square.

 
6a. 
 
 
Answer:

There are two solutions to this equation: x = -1 + 2i and x = -1 - 2i.


 
6b. 
 
 
Answer:

 

There are two solutions to this equation: x = 7/4 + 1/4 = 2 and x = 7/4 - 1/4 = 6/4 = 3/2 .


 
 
Problems 7a - 7b:  Solve the given equation by using the quadratic equation.

 
7a.
 
 
Answer:

 
7b. 
 
 
Answer:

Write in standard form:

Put in quadratic formula:


 
 
Problems 8a - 8b:  Solve the given equation by factoring.

 
8a. 
 
 
Answer:

 

There are three solutions to this equation: a = 0, a = -3, and a = 3.


 
8b. 
 
Answer:

There are three solutions to this equation: x = 5, x = -2, and x = 2.


 
 
Problems 9a - 9b:  Solve the given equation.

 
9a. 
 
 
Answer:

Checking for extraneous solutions:

Since we got a false statement, y = 3 is an extraneous solution.

This means there is no solution to this equation.


 
9b. 
 
 
Answer:

Both x = -1 and x = 3 check.

There are two solutions to this equation: x = -1 and x = 3.


 
 
Problem 10a:  Solve the given equation.

 
10a. 
 
 
Answer:

a = 8 does check.

There is one solution to this equation: a = 8.


 
 
Problems 11a - 11b:  Solve the given equation.

 
11a. 
 
 
Answer:

Writing the equation in standard form:

Substitution:

Substitute in t and solve:

 

Substitute value in for t and solve for x:

There are two solutions to this equation: x = -343 and x = 1.


 
11b. 
 
 
Answer:

Substitution:

Substitute in t and solve:

 

Substitute value in for t and solve for y:

There are two solutions to this equation: y = 3 and y = -1.


 
 
Problems 12a - 12b:  Solve the given equation.

 
12a. 
 
 
Answer:

First solution:

Second solution:

There are two solutions to this equation: x = 8/3 and x = -16/9.


 
12b. 
 
 
Answer:

Since the absolute value is set equal to a negative number, there is no solution.


 
 
Problems 13a - 13b:  Solve, write your answer in interval notation and graph the solution set.

 
13a. 
 
 
Answer:

 

Interval notation: 

Graph: 


 
13b. 
 
 
Answer:

 

Interval notation: 

Graph: 


 
 
Problems 14a - 14d:  Solve, write your answer in interval notation and graph the solution set.

 
14a. 
 
 
Answer:

Use the definition of absolute value to set up:

 

Interval notation: 

Graph: 


 
14b. 
 
 
Answer:

Use the definition of absolute value to set up:

 

Interval notation: 

Graph: 


 
14c. 
 
 
Answer:

Isolate the absolute value:

Since the absolute value is ALWAYS positive and in this problem it is set greater than or equal to a negative number, the answer is all real numbers.


 
14d. 
 
 
Answer:

Since the absolute value is ALWAYS positive and in this problem it is set less than a negative number, the answer is no solution.


 
 
Problems 15a - 15b:  Solve, write your answer in interval notation and graph the solution set.

 
15a. 
 
 
Answer:

Write in standard form:

Solve quadratic equation:

 

Mark off boundary points on number line:

Note that the two boundary points create three sections on the graph: 
.

Chose -1 in first interval to check:

Since 4 is positive and we are looking for values that cause our quadratic expression to be less than or equal to  0 (negative or 0), would not be part of the solution.
 

Chose 0 in second interval to check:

Since -3 is negative and we are looking for values that cause our expression to be less than or equal to 0 (negative or 0),  would be part of the solution.
 

Chose 4 in third interval to check:

Since 9 is positive and we are looking for values that cause our quadratic expression to be less than or equal to  0 (negative or 0), would not be part of the solution.
 

Interval notation: 

Graph: 


 
15b. 
 
 
Answer:

Set numerator equal to 0 and solve:

Set denominator equal to 0 and solve:

 

Mark off boundary points on number line:

Note that the two boundary points create three sections on the graph: , and .

Chose -4 in first interval to check:

Since 1/8 is positive and we are looking for values that cause our quadratic expression to be greater than  0 (positive), would  be part of the solution.
 

Chose 0 in second interval to check:

Since -3/4 is negative and we are looking for values that cause our quadratic expression to be greater than  0 (positive), would not be part of the solution.

Chose 5 in third interval to check:

Since 8 is positive and we are looking for values that cause our quadratic expression to be greater than  0 (positive), would  be part of the solution.
 

Interval notation: 

Graph: 


 




All contents
August 25, 2002