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
D
Orders Hi.e. powers or rootsL
Division
¸
Joint third priority
M
Multiplication
´
Joint third priority
A
Addition
+
Joint fourth priority
S
Subtraction
-
Joint fourth priority
O
43 ,
25
Second priority
2 | Algebra
ASK Week- Autumn 2012
Example Add brackets as appropriate to make these equalities true 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
5 4 - 3 + 2 = ?
ASK Week- Autumn 2012
Algebra | 3
N-th power and n-th root Exponentiation is a mathematical operation, written as an , involving two numbers, the base a and the exponent (or power or indices) n. The exponent is usually shown 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 a = an e.g. 4 4 4 4 4 = 45 n times
5 times
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 = am an
24+6 = 24 26
Han Lm = an m
I3.53 M = 3.512 4
I7 xL3 = 73 x3
(a bLm = am bm 1 an
a-n =
an am
an-m = a0 = 1 1
n
an = m
2-1 =
3-3 = HH1 3L 3L 3 = 54
56
1=
n
255 2 =
1 33
=
1 27
= 54-6 = 5-2
3m 3m
= 3m-m = 30 3
101 3 =
a
a n = am
1 2
10
255 = J 25 N = 55 5
82 3 = I81 3 M = 22 = 4 2
or = I82 M = 1 3
8-2 3 = I8 M
1 3 -2
1
3
64 = 4
= 2-2 =
1 4
We interpret a 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
n
a ”or a “surd”). Sometimes there is more than one value, e.g.
49 = ± 7 =
+7 -7
4 | Algebra
ASK Week- Autumn 2012 1
We interpret a 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
Ü
n
a ”or a “surd”). Sometimes there is more than one value, e.g.
Note: ab = aHb L ¹ Iab Mc = ab c c
c
m³0
6
n³0
8
laws, assuming a, b > 0:
a 6 IbM
=
=
a b
´
a b
6 times a a ´b b
6 times
a´a´a´a´a´a b´b´b´b´b´b 6 times
=
´
a6 b6
am
am an
´
an
a b
´
a b
Ham Ln
Ha bLm
J N
a m
b
a0
a-1
a-m
1
am
49 = ± 7 =
+7 -7
ASK Week- Autumn 2012
Algebra | 5
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 3 x3
Simplify
5x-5 y
13 x2
2x-2 y
Step 1. Factorise the numerators and denominators.
3 x1 x2
2 Hx - yL
5 Hx - yL 13 x2
Step 2. Cancel factors which are common to the numerator and denominator.
3 x1
5
2
13
Step 3. Multiply the remaining factors in the numerator and multiply the
3 x3
5x-5 y
2x-2 y
2
13 x
=
15 x 26
remaining factors in the denominator.
6 | Algebra
ASK Week- Autumn 2012
Example Evaluate the following, expressing each answer in its simplest form. x2 - 6 x + 9
1.
x2 - 8 x + 16
x2 - 16
x2 + 6 x + 9
Tick to show solution
x2 - 6 x + 9
x2 - 8 x + 16
x2 - 16 4 x2
2.
x2 + 6 x + 9
=
Hx-4L Hx-3L2
Hx+3L2 Hx+4L
x2 - 3 x - 10
x2 - 25
28 x4
Tick to show solution
4 x2
x2 - 3 x - 10
x2 - 25
=
28 x4
Dividing Fractions 4
Simplify
x ¸
x-2
2x - 4
Step 1. Invert the second fraction.
4
2x - 4 ¸
x-2
x
Step 2. Change ¸ to ´
4
2x - 4 ´
x-2
x
x+2
7 x2 Hx+5L
ASK Week- Autumn 2012
Algebra | 7
Step 3. Proceed with the multiplication of fractions.
4
2x - 4 ´
4 =
x-2
´
x
x-2
4
x ¸
x-2
= 2x - 4
2 Hx - 2L x
8 x
Example Evaluate the following, expressing each answer in its simplest form. 1.
3 Hx - 4L 2 Hx - 2L
¸
5 Hx - 4L
17 H x + 6L
Tick to show solution
3 Hx - 4L 2 Hx - 2L 2.
¸
5 Hx - 4L
17 H x + 6L
x2 - 6 x + 9 x2 - 16
=
51 Hx+6L 10 Hx-2L
x2 - 8 x + 15
¸
x2 - 6 x + 8
Tick to show solution
x2 - 6 x + 9 2
x2 - 8 x + 15 ¸
x - 16
3.
x - 6x + 8
x2 - 14 x + 40 -6 - x + x2
=
2
x2 - 6 x + 8 ¸
x2 - 4
Hx-3L Hx-2L Hx-5L Hx+4L
8 | Algebra
ASK Week- Autumn 2012
Tick to show solution
x2 - 14 x + 40
x2 - 6 x + 8 ¸
2
2
-6 - x + x
= 1-
x - 4
7 x-3
Addition and Subtraction of Fractions Express as single fraction
3 Hx + 4L
7x-1 +
15
3
Step 1. Factorise (if necessary) the denominators of the fractions.
2 Hx + 4L
7x-1 +
=
15
2 Hx + 4L
7x -1 +
5 3
3
3
Step 2. Find the lowest common denominator.
2 Hx + 4L
2 Hx + 4L
7x-1 +
=
15
5 3
3
common denominator
+
H7 x - 1L 5 3 5
common denominator
Step 3. Add or subtract the expressions on the numerators.
2 Hx + 4L 7 x - 1 + = 5 3
1 15
Example Express as single fraction 1.
4x
7z -
y-3
y-3
H37 x + 3L
ASK Week- Autumn 2012
Algebra | 9
Tick to show solution
4x
7z -
y- 3 2
= y- 3
4 x-7 z y-3
3
2.
x
y
Tick to show solution
2
3 -
=
x
2 y-3 x xy
y 1
3.
1 -
y-3
y+1
Tick to show solution
1
1 -
y- 3
4.
= y+ 1
4x
4 Hy-3L Hy+1L
7x -
y-3
y+1
Tick to show solution
4x
7x -
y- 3
y+ 1
x H3 y-25L
= - Hy-3L Hy+1L
10 | Algebra
ASK Week- Autumn 2012
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
to m.
8
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
ASK Week- Autumn 2012
Algebra | 11
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 2 M
-4
?
Tick to show solution
H5-3 2 L-4 = 15 625
6. What is
JI 2 M
N ?
1 -3 2 -2
Tick to show solution
JI 2 M
N
1 -3 2 -2
=
1 8
4
7. Simplify
x
x-2
2x - 4
Tick to show solution
4
x
x-2 2x - 4
=
2x
Hx-2L2
8. Simplify the following expression 23 x6 y2 x-1 y-2 2-5 42
12 | Algebra
ASK Week- Autumn 2012
Tick to show solution
Ix6 x-1 M Iy2 y-2 M I23 2-5 42 L 4 x5
p4 y8 p y 9. Simplify
z3
p7
Tick to show solution
p4 y8
p4 y8 p y = z3
p7
z3
py
p4 p p2
=
y9 2 3
p z
Cancel common factors
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
ASK Week- Autumn 2012
Algebra | 13
Tick to show solution
36存 H6 - 2L = 9
13. Add brackets as appropriate to make the equality true
36 存 6 - 2 = 4
Tick to show solution
No brackets needed: 36存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 - 5L if u = 3.1, v = 5.9 5u-3v-5
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.
14 | Algebra
ASK Week- Autumn 2012
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 Equations An arbitrary quadratic equation a x2 + b x + c = 0 can be solved by formula
-b Âą x1,2 =
b2 - 4 a c 2a
where D = b2 - 4 a c is the discriminant of the quadratic polynomial a x2 + 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.
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. 2012 ASK WeekAutumn 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.
Algebra | 15
16 | Algebra
ASK Week- Autumn 2012
a
2.82
b
2.5
c
-2.4
Zeros
Ă&#x2C6;
Critical Points
2.8 x2 + 2.5 x - 2.4 10
5
-2
1
-1
-5
-10
2
ASK Week- Autumn 2012
Algebra | 17
a
11
b
15
c
4
reset
The equation is:
11x2 + 15x + 4 = 0 To solve it, use the quadratic formula:
x = H-15 ±
Simplify:
x = H-15 ±
H152 - 4 ´ 11 ´ 4L L H2 ´ 11L
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:
4
x1 = - 11 x2 = - 1
18 | Algebra
ASK Week- Autumn 2012
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
ASK Week- Autumn 2012
Algebra | 19
Tick to show solution
x1 = -4 x2 = -4 x2 + 7 x - 21 = 0
5.
Tick to show solution
x1 =
1 2
x2 =
1 2
J-7 -
133 N
J 133 - 7N
Solve ax2 + b x + c = 0 , using the quadratic formula
a
3
b
-2
c
-15.
3 x2 - 2 x - 15. Â&#x2021; 0 x1 = -1.92744 x2 = 2.59411
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20 | Algebra
ASK Week- Autumn 2012
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ASK Week – Autumn 2012
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