# College Algebra First Day Quiz Solutions

1. Evaluate:

Solution

Reading the expression as “the additive inverse of three-squared” 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 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

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
point-slope 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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