MATH 1090 Algebra and Trigonometry
Problem Set 1 - Solutions

2002 Winter

Note

List all numbers from the following set that are
1. natural numbers
2. whole numbers
3. integers
4. rational numbers
5. irrational numbers
6. real numbers
, a natural number.
Any finite or recurring decimal is a rational number.
The square root of any negative number is not a real number at all.
Therefore, in this set,
The natural numbers are
The whole numbers are
The integers are
The rational numbers are
The irrational numbers are
The real numbers are

Rewrite each expression without absolute value bars.
1. | 2 - p |
  The value of p is approximately 3.14 > 2.
  Therefore | 2 - p | =
  
  p - 2
2. for each of x = -5 and x = +5

Simplify the algebraic expression 3 (4 x + y) - 5 (2 x - y)
3 (4 x + y) - 5 (2 x - y) = (12 x + 3 y) + (- 10 x + 5 y)
= (12 - 10) x + (3 + 5) y =

2 x + 8 y

Simplify the algebraic expression

The Earth is 1.5×10¹¹ metres away from the Sun, on average, (to a good approximation). Light travels 3×10⁸ metres per second, (to a good approximation). How many seconds does it take light to travel from the Sun to the Earth?

Speed = distance / time , so time = distance / speed
= 1.5×10¹¹ / 3×10⁸ = 0.5×10^{11 - 8} = 0.5×10³ = 5×10² =

500 seconds

[Thus, when we see the Sun, we see it not as it is now, but as it was 8 minutes and 20 seconds ago.]

Rationalize the denominator in

Factor, where possible, the following trinomials
1. 4 x² + 16 x + 15
  If a factorization of the form (ax + b)(cx + d) exists with integer values of a, b, c, d, then
  (a, c) = (4, 1), (2, 2) or (1, 4) only and
  (b, d) = (15, 1), (5, 3), (3, 5) or (1, 15) only.
  We also need ad + bc = 16.
  The only distinct possibility is
  
  (2 x + 3) (2 x + 5)
2. 4 x² + 16 x + 16
  This is a perfect square, of the form (ax + b)² = a²x² + 2 ab x + b² ,
  with a = 2 and b = 4.
  Therefore the factorization is
  
  (2 x + 4)²
3. 4 x² + 16 x + 17
  If a factorization of the form (ax + b)(cx + d) exists with integer values of a, b, c, d, then
  (a, c) = (4, 1), (2, 2) or (1, 4) only and
  (b, d) = (17, 1) or (1, 17) only.
  We also need ad + bc = 16.
  No combination of these values will produce the correct middle coefficient of 16.
  Therefore, using real integers, there is
  
  NO factorization.

Factor and simplify the algebraic expression (x + 4)^-1/2 - (x + 4)^-3/2
(x + 4)^-1/2 - (x + 4)^-3/2 = (x + 4)^(2-3)/2 - (x + 4)^-3/2
= ((x + 4)^2/2 - 1) (x + 4)^-3/2 =

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Dr. G.H. George