Algebra Refresher

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 = ?

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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

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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

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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

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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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