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College Algebra First Day Quiz Solutions
1. Evaluate:
Solution
Reading the expression as “the additive inverse of threesquared” obtains the
correct answer, −9 .
Note the difference that parentheses make :
2. Evaluate:
Solution
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:
Solution
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 .
Solution
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.
Solution
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 pointslope form
. This solution will use
the pointslope 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
equation.
Let P_{1} = (−2, 1) and P_{2} = (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) .
Solution
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 m_{1} = −1, the slope of any
line perpendicular to it
must have slope m_{2} =1. Substitute this value and the given point into the
pointslope equation to
obtain
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:
Solution
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.)
Solve:
Solution
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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