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MATH 1110 (Lecture 002)
August 30, 2013
Pre-Calculus Review Problems — Solutions
1 Algebra and Geometry
Problem 1. Give equations for the following lines in both point-slope and slope-intercept form.
(a) The line which passes through the point (1, 2) having slope 4.
(b) The line which passes through the points (− 1 , 1) and (2, −1).
(c) The line parallel to y = 12 x + 2, with y-intercept (0, −1).
(d) The line perpendicular to y = − 3 x + 1 which passes through the origin.
Solution: (a) The point-slope form is
y − 2 = 4(x − 1).
Solving for y,
y = 4(x − 1) + 2
= 4x − 4 + 2
= 4x − 2 ,
yields the slope-intercept form,
y = 4x − 2.
(b) First, we compute the slope using the familiar “rise-over-run” formula,
m =
The point-slope form (using the first point) is,
y − 1 = −
(x + 1),
and solving for y yields the slope-intercept form,
y = −
x +
(c) The slope of our desired line is 12 , since parallel lines must have the same slope. The point-slope form
is,
y − (−1) =
(x − 0),
and the slope-intercept form is
y =
x − 1.
(d) The slope of our desired line is
1 3 , since it must be the negative reciprocal of the slope any line to which it is perpendicular. The point-slope form is,
y − 0 =
(x − 0),
and the slope-intercept form is,
y =
x.
Problem 2. Find the point of intersection, if there is one, between the following lines:
(a) y = −x + 5 and y − 2 = 3(x + 1)
(b) The line passing through (− 1 , −2) and the origin, and the line y = 2x − 2.
Solution: (a) First, we write both lines in slope-intercept form,
y = −x + 5 y = 3x + 5.
If (x, y) is a point of intersection of the lines, it must satisfy both equations. Assuming (x, y) is as such,
we have that
−x + 5 = 3x + 5
−x = 3x
x = 0.
Thus, x = 0. To find y, we can plug x = 0 into either one of the original equations, and get that y = 5.
Thus, (0, 5) is the (unique) point of intersection.
(b) The line passing through (− 1 , −2) and the origin has slope
m =
and can be expressed by the equation y = 2x. But, this line is parallel to (and distinct from) the line
y = 2x − 2, so they cannot have any points of intersection.
Problem 3. Find all real roots x of the following polynomials, and factor into irreducible polynomials.
(a) 6x^2 + 5x + 1
(b) −x^2 + x + 1
(c) 2x^2 − 3 x + 5
(d) x^3 + 6x^2 − 7 x
(e) x^3 − x^2 + x − 1
(f) x^4 − 2 x^2 + 1
Solution: Note that a polynomial is irreducible if it cannot be factored into non-constant polynomials
with real coefficients.
(a)
6 x^2 + 5x + 1 = 6x^2 + 3x + 2x + 1
= 3x(2x + 1) + 1(2x + 1)
= (3x + 1)(2x + 1).
This factors the polynomial into irreducibles, and shows that its roots are x = − 13 and x = − 12.
(b) We use the quadratic formula:
x =
(a) 3
x = x − 4
(b)
x + 2 +
x − 2 =
4 x − 2
(c) x = 4 3
x.
(d) x x−−^12 + (^2) xx+2+1 = 0
Solution: (a) First, note that the presence of
x means that any solutions x must be ≥ 0.
x = x − 4
9 x = (x − 4)
2
9 x = x
2 − 8 x + 16
0 = x^2 − 17 x + 16
0 = (x − 16)(x − 1).
The solutions the last equation are x = 1 and x = 16, and since these are both positive, they are our
solutions.
(b) The presence of
x + 2,
x − 2 and
4 x − 2 means that any solution x must satisfy x ≥ −2, x ≥ 2,
and x ≥
1 2 , but the first and third of these are redundant, so it suffices to look for solutions with^ x^ ≥^ 2. √ x + 2 +
x − 2 =
4 x − 2
(
x + 2 +
x − 2)^2 = 4x − 2
(x + 2) + 2
x + 2
x − 2 + (x − 2) = 4x − 2
2
x + 2
x − 2 + 2x = 4x − 2
2
x + 2
x − 2 = 2x − 2
4(x + 2)(x − 2) = (2x − 2)^2
4(x
2 − 4) = 4x
2 − 8 x + 4
4 x^2 − 16 = 4x^2 − 8 x + 4
−20 = − 8 x
5
2
= x.
Note that x = 52 ≥ 2, as required, so this is the solution.
(c) Every real number has a cube root, so 3
x does not impose any restrictions on our solution. Clearly
x = 0 is a solution, so in the following derivation, we can assume that x 6 = 0.
x = 4 3
x
x^3 = 64x
x
2 = 64 (since x 6 = 0)
x = ± 8.
Thus, x = 0 and x = ±8 are the solutions.
(d) Note that any solution x cannot be equal to 2 or −2.
x − 1
x − 2
2 x + 1
x + 2
x − 1
x − 2
x + 2
x + 2
2 x + 1
x + 2
x − 2
x − 2
x^2 + x − 2
x^2 − 4
2 x^2 − 3 x − 2
x^2 − 4
3 x^2 − 2 x − 4
x^2 − 4
The only way for this equation to be true is if the numerator on the left-hand side is 0, which occurs
exactly when x is a root of 3x^2 − 2 x − 4. We use the quadratic formula,
x =
Since neither of these solutions are equal to 2 or −2, we have that these are the solutions to original
equation.
Problem 5. Find the equations of the following shapes.
(a) A circle of radius 2, centered at (1, 2).
(b) A circle centered at the origin, and tangent to the line y = − 2 x + 2.
Solution: (a) The equation of the circle is
(x − 1)
2
2 = 4.
(b) [This is trickier. If you couldn’t do this problem, that is okay!] Since the circle is centered
at the origin and tangent to y = − 2 x + 2, it must intersect y = − 2 x + 2 at the point on this line which is
nearest to the origin. This is given by the intersection of y = − 2 x + 2 with the perpendicular line y = 12 x
through the origin. We can find their intersection,
− 2 x + 2 =
x
x
= x.
Plugging this in for x in the equation y = 12 x yields y = 25 , so the point of intersection is ( 45 , 25 ). The
radius r of our circle is the distance from the origin to the point ( 45 , 25 ), and so
r
2
Thus, the equation of the circle is
x^2 + y^2 =
Thus, the solution is {x : x ≤ 5 / 4 }, or equivalently, (−∞, 5 /4].
(b) Due to the absolute value, there are two cases to consider: If 3x − 7 ≥ 0, then we have
0 ≤ 3 x − 7 < 4.
So,
0 ≤ 3 x − 7 < 4
7 ≤ 3 x < 11
7 / 3 ≤ x < 11 / 3.
Thus, the solution in this case is {x : 7/ 3 ≤ x < 11 / 3 }, or [7/ 3 , 11 /3).
If 3x − 7 < 0, then we have
0 < −(3x − 7) < 4.
So,
0 < −(3x − 7) < 4
0 > 3 x − 7 > − 4
7 > 3 x > 3
7 / 3 > x > 1.
Thus, the solution in this case is {x : 1 < x < 7 / 3 }, or (1, 7 /3).
The solution in general is the union of these two sets, namely {x : 7/ 3 ≤ x < 11 / 3 }∪{x : 1 < x < 7 / 3 },
which is just {x : 1 < x < 11 / 3 }, or (1, 11 /3).
(c) Note that (x − 1)^2 = |x − 1 |^2 , so (x − 1)^2 < 9 implies that
|x − 1 | < 3.
There are two cases to consider: If x − 1 ≥ 0, then
0 ≤ x − 1 < 3
1 ≤ x < 4.
Thus, the solution in this case is {x : 1 ≤ x < 4 }. If x − 1 < 0, then
0 < −(x − 1) < 3
0 > x − 1 > − 3
1 > x > − 2.
Thus the solution in this case is {x : − 2 < x < 1 }. Thus, the solution in general is {x : − 2 < x < 4 }, or
(− 2 , 4).
(d) The presence of
x − 1 tells us that the solution must be contained in [1, ∞), i.e., any x in the solution
set is ≥ 1.
√ x − 1 ≥ 2
x − 1 ≥ 4
x ≥ 5.
Thus, the solution is {x : x ≥ 5 }, or [5, ∞).
4 Trigonometry
Problem 9. Fill in the following table with exact values:
θ in degrees θ in radians 0 ≤ θ < 2 π sin θ cos θ tan θ
0 ◦^0 0 1
30 ◦^ π/ 6 1 / 2
45 ◦^ π/ 4 1 /
60 ◦^ π/ 3
90 ◦^ π/ 2 1 0 not defined
120 ◦^2 π/ 3
135 ◦^3 π/ 4 1 /
150 ◦^5 π/ 6 1 / 2 −
180 ◦^ π 0 − 1 0
210 ◦^7 π/ 6 − 1 / 2 −
225 ◦^5 π/ 4 − 1 /
240 ◦^4 π/ 3 −
270 ◦^3 π/ 2 − 1 0 not defined
300 ◦^5 π/ 3 −
315 ◦^7 π/ 4 − 1 /
330 ◦^11 π/ 6 − 1 / 2
Problem 10. Find the exact values for the following expressions.
(a) tan θ when θ is in the third quadrant and sin θ = 45. [Note: There was a typo here! It should
be sin θ = − 45 .]
(b) sin 12 π. (Hint: remember your trig identities?)
Solution: (a) Recall that when θ is in standard position relative to the xy-axis,
sin θ =
x 0
r
where the point (x 0 , y 0 ) is the intersection of the terminal ray of the angle with a circle centered at the
origin and having radius r. Since θ lies in the third quadrant, the x 0 must be negative, and we can take
it to be −4, and with r = 5. Since,
cos θ =
y 0
r
we must find y 0. By Pythagoras, the length y 0 is 3 (this is a 3-4-5-right triangle), and it is negative, since
θ is in the third quadrant. That is, y 0 = −3. Thus,
tan θ =
x 0
y 0
(b) First, note that π/12 = π/ 3 − π/4. We can use the following trig identity:
sin(θ − φ) = sin θ cos φ − cos θ sin φ.
