algebra refresher
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ASK weeek workshopTRANSCRIPT
Free Mathematica demonstration projects were used in this presentation ( see http : // demonstrations.wolfram.com/ )
Algebra Refresher
Order of Operations
Types of Numbers
Whole numbers are 1, 2, 3, ...
Natural numbers are 0, 1, 2, 3, ...
Integers are 0, ±1, ±2, ±3, ...
Rational numbers are numbers that cannot be expressed as the ratio of two integers e.g. Π
Decimal numbers e.g. 2.13, 5.169, 0.0132
Order of operations
B Brackets H L First priority
O Orders Hi.e. powers or rootsL 43, 25 Second priority
D Division ¸ Joint third priority
M Multiplication ´ Joint third priority
A Addition + Joint fourth priority
S Subtraction - Joint fourth priority
Example
Add brackets as appropriate to make these equalities true 5�4 - 3 + 2 = ?
1. 5� 4 - 3 + 2 = 19
Tick to show solution
No brackets needed: 5�4-3+2 =19
2. 5� 4 - 3 + 2 = 7
Tick to show solution
5� H4 - 3L + 2 = 7
3. 5� 4 - 3 + 2 = 15
Tick to show solution
5�H4-3+2L = 15
or 5�4-H3+2L = 15
4. 5� 4 - 3 + 2 = -5
Tick to show solution
5�H4-H3+2 LL = -5
2 | Algebra ASK Week- Autumn 2012
N-th power and n-th rootExponentiation is a mathematical operation, written as an , involving two numbers, the base a and the exponent (or power or indices) n. The exponent is usuallyshown as a superscript to the right of the base.
Ü Note: When n is a positive integer, exponentiation corresponds to repeated multiplication
a�a�a� ....�a�an times
= an e.g. 4�4�4�4�45 times
= 45
Suppose now that a and b are positive real numbers and that m and n are arbitrary real numbers. Then the following rules are the basic Laws of Indices:
Law Example
am+n = a
ma
n2
4+6= 242
6
HanLm= an � m I3.5
3M4= 3.5
12
(a bLm = am
bm I7 xL3= 7
3x
3
a-n =
1
an
2-1=
1
2
3-3= HH1�3L�3L�3 =
1
33
=1
27
an-m =
an
am
54
56
= 54-6= 5
-2
a0= 1 1=
3m
3m
= 3m-m = 3
0
a
1
n = an
101�3 = 10
3
a
m
n = am
n
255�2= 25
5 = J 25 N5
= 55
82�3= I81�3M2
= 22= 4
or = I82M1�3= 64
3
= 4
8-2�3= I81�3M-2
= 2-2 =
1
4
We interpret a1
n to mean a number which gives the value a when it is raised to the power n.
It is called an “n-th root of a” (or a “radical an
”or a “surd”). Sometimes there is more than one value, e.g. 49 = ±7 =+7
-7
ASK Week- Autumn 2012 Algebra | 3
We interpret a1
n to mean a number which gives the value a when it is raised to the power n.
It is called an “n-th root of a” (or a “radical an
”or a “surd”). Sometimes there is more than one value, e.g. 49 = ±7 =+7
-7
Ü Note: abc= aHbcL ¹ IabMc
= ab c
m ³ 0 6
n ³ 0 8
laws, assuming a, b > 0: am
an a
m
an
HamLn Ha bLm J a
bNm
a0
a-1
a-m
a
1
m
I a
bM6
=a
b´
a
b´
a
b´
a
b´
a
b´
a
b
6 times
=a ´ a ´ a ´ a ´ a ´ a
6 times
b ´ b ´ b ´ b ´ b ´ b6 times
=a6
b6
4 | Algebra ASK Week- Autumn 2012
Rearranging and simplifying formulae� Factorisation
Factorise 3 x + 24 y.
Step 1. Factorise each term if possible
3 x + 24 y = 3� x + 3�8� y
Step 2. Take the common factor out
3� x + 3�8� y = 3 Hx + 8 yL
3 x + 24 y = 3 Hx + 8 yL
� Multiplying Fractions
Simplify 3 x3
2 x - 2 y�
5 x - 5 y
13 x2
Step 1. Factorise the numerators and denominators.
3 x1
x2
2 Hx - yL�
5 Hx - yL13 x
2
Step 2. Cancel factors which are common to the numerator and denominator.
3 x1
2
�5
13
Step 3. Multiply the remaining factors in the numerator and multiply the remaining factors in the denominator.
3 x3
2 x - 2 y�
5 x - 5 y
13 x2=
15 x
26
ASK Week- Autumn 2012 Algebra | 5
Example
Evaluate the following, expressing each answer in its simplest form.
1. x2 - 6 x + 9
x2 - 16�
x2 - 8 x + 16
x2 + 6 x + 9
Tick to show solution
x2 - 6 x + 9
x2 - 16�
x2 - 8 x + 16
x2 + 6 x + 9=
Hx-4L Hx-3L2
Hx+3L2 Hx+4L
2. 4 x2
x2 - 25�
x2 - 3 x - 10
28 x4
Tick to show solution
4 x2
x2 - 25�
x2 - 3 x - 10
28 x4=
x+2
7 x2 Hx+5L
� Dividing Fractions
Simplify 4
x - 2¸
x
2 x - 4
Step 1. Invert the second fraction.
4
x - 2¸
2 x - 4
x
Step 2. Change ¸ to ´
4
x - 2´
2 x - 4
x
6 | Algebra ASK Week- Autumn 2012
Step 3. Proceed with the multiplication of fractions.
4
x - 2´
2 x - 4
x=
4
x - 2´
2 Hx - 2Lx
4
x - 2¸
x
2 x - 4=
8
x
Example
Evaluate the following, expressing each answer in its simplest form.
1. 3 Hx - 4L2 Hx - 2L
¸5 Hx - 4L
17 H x + 6L
Tick to show solution
3 Hx - 4L2 Hx - 2L
¸5 Hx - 4L
17 H x + 6L=
51 Hx+6L10 Hx-2L
2. x2 - 6 x + 9
x2 - 16¸
x2 - 8 x + 15
x2 - 6 x + 8
Tick to show solution
x2 - 6 x + 9
x2 - 16¸
x2 - 8 x + 15
x2 - 6 x + 8=
Hx-3L Hx-2LHx-5L Hx+4L
3. x2 - 14 x + 40
-6 - x + x2¸
x2 - 6 x + 8
x2 - 4
ASK Week- Autumn 2012 Algebra | 7
Tick to show solution
x2 - 14 x + 40
-6 - x + x2¸
x2 - 6 x + 8
x2 - 4= 1 -
7
x-3
� Addition and Subtraction of Fractions
Express as single fraction 3 Hx + 4L
15+
7 x - 1
3
Step 1. Factorise (if necessary) the denominators of the fractions.
2 Hx + 4L15
+7 x - 1
3
=2 Hx + 4L
5� 3
+7 x - 1
3
Step 2. Find the lowest common denominator.
2 Hx + 4L15
+7 x - 1
3
=2 Hx + 4L
5� 3
common denominator
+H7 x - 1L � 5
3� 5
common denominator
Step 3. Add or subtract the expressions on the numerators.
2 Hx + 4L5
+7 x - 1
3=
1
15H37 x + 3L
Example
Express as single fraction
1. 4 x
y - 3-
7 z
y - 3
8 | Algebra ASK Week- Autumn 2012
Tick to show solution
4 x
y - 3-
7 z
y - 3=
4 x-7 z
y-3
2. 2
x
-3
y
Tick to show solution
2
x
-3
y
=2 y-3 x
x y
3. 1
y - 3-
1
y + 1
Tick to show solution
1
y - 3-
1
y + 1=
4
Hy-3L Hy+1L
4. 4 x
y - 3-
7 x
y + 1
Tick to show solution
4 x
y - 3-
7 x
y + 1= -
x H3 y-25LHy-3L Hy+1L
ASK Week- Autumn 2012 Algebra | 9
Exercises
1. Change the subject of the formula x = y + k z to z.
Tick to show solution
x - y = kz �
z =x- y
k
2. Change the subject of the formula 5 r =5 z H1 + mL
8 to m.
Tick to show solution
40 r = 5 z H1 + mL �
40 r - 5 z = 5 z m � 8 r - z = z m
m =8 r-z
z
3. Simplify the following expression 4 + 9 Ht - 5L + t.
Tick to show solution
4 + 9 t - 45 + t �
H4 - 45L + H9 t + tL �
� 10 t - 41
4. Simplify the following expression (2 - nL Hn - 7L + 6 n - 3
10 | Algebra ASK Week- Autumn 2012
Tick to show solution
2 n - n2 - 14 + 7 n + 6 n - 3 �
-n2 + H2 n + 7 n + 6 nL - 14 - 3 �
H2 - nL Hn - 7L + 6 n - 3 = -n2 + 15 n - 17
5. What is I5-3�2M-4 ?
Tick to show solution
H5-3�2L-4 = 15 625
6. What is JI 1
2M-3�2N
-2
?
Tick to show solution
JI 1
2M-3�2N
-2=
1
8
7. Simplify 4
x - 2�
x
2 x - 4
Tick to show solution
4
x - 2�
x
2 x - 4=
2 x
Hx-2L2
8. Simplify the following expression 23 x6 y2 x-1 y-2 2-5 42
ASK Week- Autumn 2012 Algebra | 11
Tick to show solution
Ix6 x-1M Iy2 y-2M I23 2-5 42L
� 4 x5
9. Simplify p
4y
8
z3
p y
p7
Tick to show solution
p4
y8
z3
p y
p7
=p4 y8
z3�
p y
p4 p p2=
Cancel common factors
y9
p2 z3
10. Add brackets as appropriate to make the equality true 5´ 7 - 3 = 20
Tick to show solution
5 ´ H7 - 3L = 20
11. Add brackets as appropriate to make the equality true 5´ 7 - 3 = 32
Tick to show solution
No brackets needed: 5 ´ 7 - 3 = 32
12. Add brackets as appropriate to make the equality true 36 ¸ 6 - 2 = 9
12 | Algebra ASK Week- Autumn 2012
Tick to show solution
36¸ H6 - 2L = 9
13. Add brackets as appropriate to make the equality true 36 ¸ 6 - 2 = 4
Tick to show solution
No brackets needed: 36¸6 - 2 = 4
14. Expand the expression 5 Hx - 3L Hy - 5L
Tick to show solution
5Hx-3LHy-5L = 5 x y - 25 x - 15 y + 75
15. Evaluate these expressions 5 Hu - 3L Hv - 5L5 u - 3 v - 5
if u = 3.1, v = 5.9
Tick to show solution
5Hu - 3LHv - 5L = 0.45
5u - 3v - 5 = -7.2
16. Evaluate the expression 5 u2
+ 4 Hv - 5L - 4 H3 u - 2L if u = 3., v = 2.
ASK Week- Autumn 2012 Algebra | 13
Tick to show solution
5u2
+ 4Hv - 5L - 4H3u - 2L = 5
17. Factorise y v - 5 y z
Tick to show solution
y v - 5 yz = y Hv - 5 zL
18. Factorise 2 Hy - 1L z + 4 Hy - 1L t
Tick to show solution
2 Hy - 1L z + 4 Hy - 1L t = 2 Hy - 1L H2 t + zL
Solution of Quadratic EquationsAn arbitrary quadratic equation a x
2 + b x + c = 0 can be solved by formula
x1,2 =- b ± b
2 - 4 a c
2 a
where D = b2 - 4 a c is the discriminant of the quadratic polynomial a x
2 + b x + c.
There are three important cases of quadratics depending on where the graph crosses the x-axis (which are roots or zeros of the equation).
Case 1.
The discriminant is strictly positive D > 0 .
The quadratic equation has two distinct, real roots.
Case 2.
The discriminant is strictly negative D < 0 .
The quadratic equation has no real roots.
Case 3.
The discriminant equals to zero D = 0 .
The quadratic equation has a pair coincident, real roots.
14 | Algebra ASK Week- Autumn 2012
There are three important cases of quadratics depending on where the graph crosses the x-axis (which are roots or zeros of the equation).
Case 1.
The discriminant is strictly positive D > 0 .
The quadratic equation has two distinct, real roots.
Case 2.
The discriminant is strictly negative D < 0 .
The quadratic equation has no real roots.
Case 3.
The discriminant equals to zero D = 0 .
The quadratic equation has a pair coincident, real roots.
ASK Week- Autumn 2012 Algebra | 15
a 2.82
b 2.5
c -2.4
Zeros È Critical Points
-2 -1 1 2
-10
-5
5
10
2.8 x2 + 2.5 x - 2.4
16 | Algebra ASK Week- Autumn 2012
a 11
b 15
c 4
reset
The equation is:
11x2 + 15x + 4 = 0
To solve it, use the quadratic formula:
x = H-15 ± H152 - 4 ´ 11 ´ 4L L � H2 ´ 11LSimplify:
x = H-15 ± 49 L � 22
The equation has two real solutions:
x = H-15 ± 7L � 22
or:
x1 = H-15 + 7L � 22
x2 = H-15 - 7L � 22
Simplify:
x1 = -4
11
x2 = -1
ASK Week- Autumn 2012 Algebra | 17
Exercises
Solve the following quadratic equations.
1. x2 - 9 x + 20 = 0
Tick to show solution
x1 = 4
x2 = 5
2. x2 + 6 x + 8 = 0
Tick to show solution
x1 = -4
x2 = -2
3. x2 + x - 12 = 0
Tick to show solution
x1 = -4
x2 = 3
4. x2 + 8 x + 16 = 0
18 | Algebra ASK Week- Autumn 2012
Tick to show solution
x1 = -4
x2 = -4
5. x2 + 7 x - 21 = 0
Tick to show solution
x1 =1
2J-7 - 133 N
x2 =1
2J 133 - 7N
Solve ax2 + b x + c = 0 , using the quadratic formula
a 3
b -2
c -15.
3 x2 - 2 x - 15. � 0
x1 = -1.92744
x2 = 2.59411
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ASK Week- Autumn 2012 Algebra | 19
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20 | Algebra ASK Week- Autumn 2012