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Algebra Self-Quiz Solutions

 

Item 1

Simplify the following expression by removing parentheses and combining like terms:

Solution 1

First we remove parentheses by distributing (careful with that negative sign in the second set) to get,

Then we combine like terms and order the terms by descending powers of x to get,

If you had difficulty with this exercise, review the tutorial Simplifying Expressions.

 

Item 2

Solve this equation for the variable x.

Solution 2

 First we remove the parentheses on the left hand side by distributing the 3 to give us,

Then we collect all the x-terms on the left side by subtracting x from both sides and collect all the constant terms on the right side by adding 3 to both sides yielding,

Finally, we divide both sides by 2 giving us the solution,

If you had difficulty with this exercise, review the tutorial Solving Linear Equations.

 

Item 3

Solve this equation for the variable x using any method you choose.

Solution 3

This is a quadratic equation, so we have several ways to solve for x.  We shall attempt to solve this equation by factoring, before trying another method if necessary.  Since the first term is , we know that the first term in each binomial factor must be x, so we begin by writing,

We proceed by noting that the two constant terms in our binomials must sum to negative 7 and multiply to positive 12.  The only way for this to happen is for both constant terms to be negative and in particular, they must be the constants -3 and -4.  This gives us the factorization shown here:

Since the product of the two binomials equals zero, then either or both must equal zero and we have the solutions,

and

Note that this equation could have been solved using the method of completing the square or by using the quadratic formula.

If you had difficulty with this exercise, review the tutorials Solving Quadratic Equations by Factoring, Solving Quadratic Equations by Completing the Square or Solving Quadratic Equations using the Quadratic Formula .

 

Item 4

Solve this inequality:

Solution 4

We treat the inequality sign much like we treat an equals sign when solving an equation and begin by collecting all the x-terms on the left hand side by subtracting 2x from both sides and collecting all the constant terms on the right hand side by subtracting 1 from both sides giving us,

Then we divide both sides by -3.  However, we must be careful here and remember that when we divide an inequality by a negative number, the sign changes direction.  So we get the solution,

If you had difficulty with this exercise, review the tutorial Inequalities.

 

Item 5

Factor this polynomial completely given that is one of the roots.

Solution 5

Saying that is one of the roots is the same as saying that  divides.  Thus, we can carry out long division of polynomials knowing that the remainder must be zero.  Here is that long division:


Therefore, we may conclude that,

So, now we must work on factoring the remaining quadratic on the right hand side.  You may notice that this is the same quadratic that we factored in problem (3).

Thus, the complete factorization is as follows:

If you had difficulty with this exercise, review the tutorials Factoring General Polynomials and Long Division of Polynomials.

 

Item 6

Expand this product using the F.O.I.L. method for multiplying binomials.


 

Solution 6

 First we write down the terms which must be included in the product.

F:
O:
I:
L:

Next combine the four terms like this,

Removing parentheses give us,

Now we notice that the two terms in the middle are like terms, so we may combine them and get the simplified form below.

So we may conclude .

If you had difficulty with this exercise, review the tutorial Multiplying Binomials (F.O.I.L. Method).

 

Item 7

Write an equation that is equivalent to the following statement:

“One less than Quentin’s age is six more than twice Paul’s age.”

Solution 7

The first thing to do when translating words into mathematics is to determine what the unknowns are and assign these quantities variables.  In this case, the unknowns are Quentin’s age, which we will represent with the variable q, and Paul’s age, which we will represent with the variable p.  The portion of the statement that says, “One less than Quentin’s age…” can be translated into the expression,

The portion of the statement that says, “…six more than twice Paul’s age,” can be translated into the expression,

Since the claim of the statement is that these two expressions have the same value, we may place an equals sign between them and generate the equivalent mathematical statement,

Now if one were to tell us either Quentin’s age or Paul’s age, we could solve this equation for the other person’s age.

If you had difficulty with this exercise, review the tutorial Translating Words into Mathematics.

 

Item 8

Find the domain and range of the function .

Solution 8

 We begin by determining the domain.  We ask ourselves the question, “Are there any values for x which make the function undefined?”  In this case, the answer is yes since the square root function is undefined whenever the number under the radical is negative.  Thus, the domain is all non-negative numbers and we write,

Domain:  {x | x ≥ 0 }.

(it is important that we have a greater than or equal to sign since we do allow x to be equal to zero).

As for the range, a look at the graph is helpful here, but we can also determine the range analytically.  Note that the square root function always takes on non-negative values.  We can see that it is zero when and increases without bound as x increases.  Therefore, the range is all non-negative numbers and we write,

Range: {y | y ≥ 0 }.

If you had difficulty with this exercise, review the tutorial Domain and Range.