Algebra Refresher

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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


5� H4 - 3L + 2 = 7

3. 5� 4 - 3 + 2 = 15


5�H4-3+2L = 15

or 5�4-H3+2L = 15

4. 5� 4 - 3 + 2 = -5


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


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


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


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


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


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


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


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



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



4 x

y - 3-

7 z

y - 3=

4 x-7 z

y-3

2. 2

x

-3

y


2

x

-3

y

=2 y-3 x

x y

3. 1

y - 3-

1

y + 1


1

y - 3-

1

y + 1=

4

Hy-3L Hy+1L

4. 4 x

y - 3-

7 x

y + 1


4 x

y - 3-

7 x

y + 1= -

x H3 y-25LHy-3L Hy+1L


Exercises

1. Change the subject of the formula x = y + k z to z.


x - y = kz �

z =x- y

k

2. Change the subject of the formula 5 r =5 z H1 + mL

8 to m.


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.


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



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 ?


H5-3�2L-4 = 15 625

6. What is JI 1

2M-3�2N

-2

?


JI 1

2M-3�2N

-2=

1

8

7. Simplify 4

x - 2�

x

2 x - 4


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



Ix6 x-1M Iy2 y-2M I23 2-5 42L

� 4 x5

9. Simplify p

4y

8

z3

p y

p7


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


5 ´ H7 - 3L = 20

11. Add brackets as appropriate to make the equality true 5´ 7 - 3 = 32


No brackets needed: 5 ´ 7 - 3 = 32

12. Add brackets as appropriate to make the equality true 36 ¸ 6 - 2 = 9



36¸ H6 - 2L = 9

13. Add brackets as appropriate to make the equality true 36 ¸ 6 - 2 = 4


No brackets needed: 36¸6 - 2 = 4

14. Expand the expression 5 Hx - 3L Hy - 5L


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


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.



5u2

+ 4Hv - 5L - 4H3u - 2L = 5

17. Factorise y v - 5 y z


y v - 5 yz = y Hv - 5 zL

18. Factorise 2 Hy - 1L z + 4 Hy - 1L t


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.


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.


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


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


Exercises

Solve the following quadratic equations.

1. x2 - 9 x + 20 = 0


x1 = 4

x2 = 5

2. x2 + 6 x + 8 = 0


x1 = -4

x2 = -2

3. x2 + x - 12 = 0


x1 = -4

x2 = 3

4. x2 + 8 x + 16 = 0



x1 = -4

x2 = -4

5. x2 + 7 x - 21 = 0


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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Algebra Refresher