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# The A to Z of mathematics a basic guide - Sidebotham T.H.

Sidebotham T.H. The A to Z of mathematics a basic guide - Wiley publishing , 2002. - 489 p.
ISBN 0-471-15045-2
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Two events are complementary if the two following conditions are both true:
1. The two events are mutually exclusive, which means both cannot happen at the same time.
2. Together the two events make up the whole sample space for the experiment of rolling a die. This means that they are the only two possible events that can take place.
In our example, rolling an odd number and rolling an even number are complementary events, because they are mutually exclusive and they make up the whole sample space, since every number rolled must be odd or even. This means that E and O are complementary events. However, 5 and O are not complementary events. They meet the first requirement to be mutually exclusive, but they do not make up the whole sample space, because they do not include the numbers 2 and 4.
Two events are independent when the outcome of one event has no effect on the outcome of the other event. Independent events usually take place when one event follows another, as in the following example. Liz has a bag that holds three red, one green, and two blue marbles. She draws one marble from the bag, has a look at its color, and replaces it back in the bag. After shaking the bag to ensure the marbles are thoroughly mixed, she then draws another marble from the bag. There are two events here:
♦ Drawing the first marble, which we call event F.
♦ Drawing another marble, which we call event A.
COMPLETING THE SQUARE 93
These two events are independent, because the outcome of F does not affect the outcome of A, and vice versa. For example, whether the first marble is red, green, or blue has no effect on the outcome of the next draw, because it is replaced in the bag before the next draw takes place. On the other hand, if the first marble is not replaced, then its outcome does affect the second draw, because there are less of that color to select on the next draw, and then F and A are not independent events.
References: Dice, Event, Probability of an Event.
COMPLETING THE SQUARE
Completing the square is something that is done to quadratic expressions to make them perfect squares. Quadratic expressions can be graphed more easily when they are written as perfect squares. Also, completing the square is a method of solving quadratic equations, especially those which do not have factors.
A quadratic expression, like x2 — 8.x + 16, is a perfect square when it is written as (x — 4)2. Other examples of perfect squares are
♦ (x + l)2, which is equal to x2 + 2x + 1
♦ (x — 3)2, which is equal to x2 — 6x + 9
♦ (x + 6)2, which is equal to x2 + 12x + 36
The quick rule for recognizing perfect squares is
Sometimes a number needs to be added to a quadratic expression to make it into a perfect square.
Example 1. What must be added tox2 + lOx + 20 to make it a perfect square, and what is the perfect square?
Solution. Write
b = 10 Matching x2 + 10a: + 20 with the formula x2 + bx + c
x2 + bx + c is a perfect square if c
Substituting b = 10 into the formula c
2
c = 25
52 = 25
94 COMPLETING THE SQUARE
x2 + 1 Ox + 25 is a perfect square, so 5 needs to be added to x2 + lOx + 20 to make it a perfect square. The perfect square is (x + 5)2. By adding 5 to x2 + lOx + 20, we have completed the square on x2 + lOx + 20.
Example 2. Write x2 — 6x + 2 in the form of (x — p)2 + q.
Solution. Write
Matching x2 — 6x + 2 with the formula
x2 + bx + c
^ . Substituting b = —6 into the formula c -
c = 9 (-3)2 = 9
6x + 2 = x2 — 6x + 9 — 7 Since x2 — 6x + 9 is a perfect square
= (x — 3)2 — 7 Replacing x2 — 6x + 9 by (x — 3)2
Completing the square is also used to solve quadratic equations, as shown in the following example.
Example 3. Solve the quadratic equation x2 — 6x + 3 = 0 by completing the square, leaving the answers to 2 dp.
Solution. Write
x2 — 6x + 3 = 0
x2 — 6x = — 3 Subtracting 3 from both sides
of the equation
x2 — 6x + 9 = —3 + 9 Adding 9 to both sides of
equation, which completes the square on the left side
x2 — 6x + 9 = 6
(x — 3)2 = 6 Replacing, x2 — 6x + 9 by
(x - 3 f
x — 3 = ±V6 Taking the square root, and
remembering the ±
COMPONENTS OF A VECTOR 95
x — 3 = -\/б or x — 3 = —\/б Writing as two answers
ж = V6 + 3 or = —ч/б + 3 Adding 3 to both sides of the
equations
x = 5.45 or x = 0.55 to 2 dp Using a calculator