College Algebra First Day Quiz Solutions

1. Evaluate:


Reading the expression as “the additive inverse of three-squared” obtains the correct answer, −9 .

Note the difference that parentheses make:

2. Evaluate:


As for the previous problem , reading the expression as “the additive inverse of the absolute value of
negative 15” obtains the correct answer, −15 .

3. Simplify:


If one used the erroneous identity , then one would erroneously conclude that
. Recall that the principal square root is never negative . Using the correct identity
, one would correctly conclude that .

Note that if the restriction x ≥ 0 is made then is correct.

4. Write a literal equation that defines subtraction .


Recall that a literal equation (or expression) is one that contains variables (letters).

a − b = a + (−b)

Example: −7 − 7 = −7 + (−7) = −14

5. What is the geometric meaning of ?

Answer: The distance from the origin (zero) to the graph of x on the number line

6. What is the geometric meaning of ?

Answer: The distance between the graphs of a and b on the number line

7. Find the equation of the line that passes through (−2, 1) and (1, − 2) . Write your answer in
standard form.


Standard form of the equation of a line is Ax + By = C . This form can be obtained from either the
slope-intercept form y = mx + b or the point- slope form . This solution will use
the point-slope form as the initial equation.

We are given points on the line, but we are not given the slope. Find the slope using the slope

Let P1 = (−2, 1) and P2 = (1, − 2) . (Recall that this is an arbitrary choice. The opposite choice could
have been made.) Then

Now let either of the given points be the point , and substitute the point and the computed
value for the slope into the equation . Letting , we obtain

y − (−2) = −(x −1) .

Rearrange this equation to obtain the standard form.

x + y = −1

8. Find the equation of the line that is perpendicular to the line in the previous problem and also
passes through (1, − 2) .


Recall that and are equations of two lines. The following apply.

The lines intersect at a unique point.
The lines are parallel.
The two lines are the same line; i.e., they are coincident.
The lines are perpendicular.

Because the line in the previous problem has slope m1 = −1, the slope of any line perpendicular to it
must have slope m2 =1. Substitute this value and the given point into the point-slope equation to

y − (−2) =1(x −1) .

The equivalent slope -intercept equation is

y = x − 3 ,

and an equivalent general form equation is

x − y = 3 .

Note that −x + y = −3 is also equivalent general form equation.

9. Simplify:


The given expression is a complex fraction; i.e., it is a fraction that contains fractions. One way to
simplify it (the way I recommend) is to clear the fractions in the numerator and denominator of the
main fraction by multiplying both by the LCD of all the fractions that appear in the numerator and
denominator .

Note that if we had been asked to provide the restrictions for the identity, we would
state that the identity holds for all real values of x not equal to 0 or −1/ 2 . ( Zero causes the given
fraction to be undefined. −1/ 2 causes the simplified fraction to be undefined.)



This is a quadratic equation. Express it in the form .

The Quadratic Formula can always be used to solve any quadratic equation. Let
us recall some properties of the discriminant , the radicand in the Quadratic Formula.

D > 0 => There are two real solutions.
D = 0 => There is one real solution, of multiplicity two.
D < 0 => There are two complex conjugate solutions.

Another use of the discriminant is that if D is a perfect square, the trinomial can be
factored over the integers, i.e., using integer coefficients .

Substituting a = 2, b = −1 and c =1 into the Quadratic Formula obtains

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