2nd term - ies jovellanos

Mathematics

3º ESO

2nd term

María Isabel Muñoz Molina

Juan Francisco Antona Blázquez

Real Instituto de Jovellanos

2nd TERM

Equations (Cont.)

Geometry

Sequences

Equations

(Continuation)

• Simultaneous equations

• Quadratic equations

Simultaneous equations

1. Linear simultaneous equations

Linear simultaneous equations are two equations containing two unknowns.

When we solve these problems we are trying to find the solution which is common to both equations.

Notice that in

{

x+ y = 9

2x+ 3y = 21, if x = 6 and y = 3 then:

• x+ y = 6+ 3 = 9 X i.e., the equation is satisfied• 2x+ 3y = 2 · 6 + 3 · 3 = 12 + 9 = 21 X i.e., the equation is satisfied.

So, x = 6 and y = 3 is the solution to the simultaneous equations

{

x+ y = 9

2x+ 3y = 21.

The solutions to linear simultaneous equations can be found by trial and error (a little tedious) orgraphically (which can be inaccurate if solutions are not integers).

However, because of the limitation of these methods, other algebraic methods are used.

2. Solution by substitution

The method of solution by substitution is used when at least one equation is given with either x ory as the subject of the formula.

Example 1

Solve simultaneously, by substitution:y = 9− x

2x+ 3y = 21

y = 9− x ......(1)2x+ 3y = 21 ......(2)

Since y = 9− x, then 2x+ 3(9− x) = 21 (substituting (1) into (2))

∴ 2x+ 27− 3x = 21

∴ 27− x = 21

∴ x = 6

and so, when x = 6, y = 9− 6 (substituting x = 6 into (1))

∴ y = 3.

Solution is: x = 6, y = 3.

Check: (1) 3 = 9− 6 X (2) 2 · 6 + 3 · 3 = 12 + 9 = 21 X

1

2 Simultaneous equations - 3 o ESO

Example 2

Solve simultaneously, by substitution:2y − x = 2x = 1 + 8y

2y − x = 2 ......(1)x = 1 + 8y ......(2)

Substituting (2) into (1) gives

∴ 2y − (1 + 8y) = 2

∴ 2y − 1− 8y = 2

∴ −6y − 1 = 2

∴ −6y = 3

∴ y = −1

2

Substituting y = −1

2into (2) gives x = 1 + 8 ·

(

−1

2

)

= −3.

The solution is: x = −3, y = −1

2.

Check: (1) 2(

−1

2

)

− (−3) = 2 X (2) 1 + 8(

−1

2

)

= −3 X

There are infinitely many points (x, y) which satisfy the first equation. Likewise, there are infinitelymany that satisfy the second. However, only one point satisfies both equations at the same time.

Exercises - Set A

1. Solve simultaneously, using substitution:

a)x = 8− 2y2x+ 3y = 13

b)y = 4 + x

5x− 3y = 0c)

x = −10− 2y3y − 2x = −22

d)x = −1 + 2yx = 9− 2y

e)3x− 2y = 8x = 3y + 12

f)x+ 2y = 8y = 7− 2x

2. Use the substitution method to solve simultaneously:

a)x = −1− 2y2x− 3y = 12

b)y = 3− 2xy = 3x+ 1

c)x = 3y − 95x+ 2y = 23

d)y = 5x7x− 2y = 3

e)x = −2− 3y3x− 2y = −17

f)3x− 5y = 26y = 4x− 12

3. Try to solve by substitution

{

y = 3x+ 1y = 3x+ 4

. What is the simultaneous solution?

4. Try to solve by substitution

{

y = 3x+ 12y = 6x+ 2

. How many simultaneous solutions do the

equations have?

Dpto. M atemáticas. IES Jovellanos. 2012

Simultaneous equations - 3 o ESO 3

3. Solution by elimination

In many problems which require the simultaneous solution of linear equations, each equation willbe of the form ax + by = c. Solution by substitution is often tedious in such situations and themethod of elimination of one of the variables is preferred.

In the method of elimination, we eliminate (remove) one of the variables by making the coefficientsof x (or y) the same size but opposite in sign and then adding the equations. This has the effectof eliminating one of the variables.

Example 3

Solve simultaneously, by elimination:4x+ 3y = 2 ......(1)x− 3y = 8 ......(2)

We sum the LHS’s and the RHS’s to get an equation which contains x only.

4x + 3y = 2+ x − 3y = 8

5x = 10

∴ x = 2

Let x = 2 in (1) ∴ 4 · 2 + 3y = 2

∴ 8 + 3y = 2

∴ 3y = −6

∴ y = −2

i.e., x = 2 and y = −2

Check: in (2): 2− 3(−2) = 2 + 6 = 8 X

The method of elimination uses the fact that:

If a = b and c = d then a+ c = b+ d.

Exercises - Set B

1. What equation results when the following are added vertically?:

a)5x+ 3y = 12x− 3y = −6

b)2x+ 5y = −4−2x− 6y = 12

c)4x− 6y = 9x+ 6y = −2

d)12x+ 15y = 33−18x− 15y = −63

e)5x+ 6y = 12−5x+ 2y = −8

f)−7x+ y = −57x− 3y = −11

2. Solve the following using the method of elimination:

a)2x+ y = 33x− y = 7

b)4x+ 3y = 76x− 3y = −27

c)2x+ 5y = 16−2x− 7y = −20

d)3x+ 5y = −11−3x− 2y = 8

e)4x− 7y = 413x+ 7y = −6

f)−4x+ 3y = −254x− 5y = 31

In problems where the coefficients of x (or y) are not the same size or opposite in sign, we may haveto multiply each equation by a number to enable us to eliminate one variable.



Example 4

Solve simultaneously, by elimination:3x+ 2y = 7 ......(1)2x− 5y = 11 ......(2)

We can eliminate y by multiplying (1) by 5 and (2) by 2.

15x + 10y = 35+ 4x − 10y = 22

19x = 57

∴ x = 3

Substituting x = 3 into equation (1) gives

3 · 3 + 2y = 7

∴ 9 + 2y = 7

∴ 2y = −2

∴ y = −1

So, the solution is: x = 3, y = −1.

Check: 3·3+2(−1) = 7 X 2·3−5(−1) = 11 X

There is always a choice whether to eliminate x or y, so our choice depends on which variable iseasier to eliminate.

Example 5

Solve by elimination:3x+ 4y = 14 ......(1)4x+ 5y = 17 ......(2)

To eliminate x, multiply both sides of (1) by 4, and (2) by −3.

12x + 16y = 56+ − 12x − 15y = −51

y = 5

Substituting y = 5 into (2) gives

4x+ 5 · 5 = 17

∴ 4x+ 25 = 17

∴ 4x = −8

∴ x = −2

Thus x = −2, y = 5.

Check: 3(−2)+4·5 = 14 X 4(−2)+5·5 = 17 X

Try now to solve this example by multiplying (1) by 5 and (2) by −4. This eliminates y ratherthan x. The final solution should be the same.

Exercises - Set C

1. Give the equation that results when both sides of the equation

a) 3x+ 4y = 2 are multiplied by 3 b) x− 4y = 7 are multiplied by −2

c) 5x− y = −3 are multiplied by 5 d) 7x+ 3y = −4 are multiplied by −3

e) −2x− 5y = 1 are multiplied by −4 f) 3x− y = −1 are multiplied by −1



2. Solve the following using the method of elimination:

a)4x− 3y = 6−2x+ 5y = 4

b)2x− y = 9x+ 4y = 36

c)3x+ 4y = 6x− 3y = −11

d)2x+ 3y = 73x− 2y = 4

e)4x− 3y = 66x+ 7y = 32

f)7x− 3y = 293x+ 4y + 14 = 0

g)2x+ 5y = 203x+ 2y = 19

h)3x− 2y = 104x+ 3y = 19

i)3x+ 4y + 11 = 05x+ 6y + 7 = 0

3. Use the method of elimination to attempt to solve:

a)3x+ y = 86x+ 2y = 16

b)2x+ 5y = 84x+ 10y = −1

Comment?

4. Problem solving

Many problems can be described mathematically by a pair of linear equations, i.e., two equations ofthe form ax+ by = c, where x and y are the two variables (unknowns).

Once the equations are formed, they can be solved simultaneously and the original problem can besolved. The following method is recommended:

Step 1: Read the problem carefully.

Step 2: Decide on the two unknowns; call them x and y, say. Do no forget the units.

Step 3: Write down two equations connecting x and y.

Step 4: Solve the equations simultaneously.

Step 5: Check your solutions with the original data given.

Step 6: Give your answer in sentence form.

The form of the original equations will help you decide whether to use the substitution method, orthe elimination method.

Example 6

Two numbers have a sum of 45 and a difference of 13. Find the numbers.

Let x and y be the unknown numbers, where x > y. Then:

x + y = 45x − y = 13

2x = 58

∴ x = 29

Substituting into the first equation

29 + y = 45

∴ y = 16

The numbers are 29 and 16.

Check: 29 + 16 = 45 X 29− 16 = 13 X



Exercises - Set D

1. The sum of two numbers is 47 and their difference is 14. Find the numbers.

2. Find two numbers with sum 28 and half their difference 2.

3. The larger of two numbers is four times the smaller, and their sum is 85. Find the twonumbers.

Example 7

When shopping in the West Indies, 5 oranges and 14 bananas cost me $1.30, and 8 oranges and9 bananas cost $1.41. Find the cost of each orange and each banana.

Let each orange cost x cents and each banana cost y cents.

∴ 5x+ 14y = 130 .....(1)8x+ 9y = 141 .....(2)

(Note: Units must be the same on both sides of each equation i.e., cents.)

To eliminate x, we multiply (1) by 8 and (2) by −5.

40x + 112y = 1040−40x − 45y = −705

67y = 335

∴ y = 5

Substituting in (2)

8x+ 9 · 5 = 141

∴ 8x = 96

∴ x = 12

Check: 5 · 12 + 14 · 5 = 130 X 8 · 12 + 9 · 5 = 141 X

Thus oranges cost 12 cents, bananas cost 5 cents each.

Exercises - Set E

1. Five pencils and 6 biros cost a total of $4.64, whereas 7 pencils and 3 biros cost a total of$3.58. Find the cost of each item.

2. Seven toffees and three chocolates cost a total of $1.68, whereas four toffees and fivechocolates cost a total of $1.65. Find the cost of each of the sweets.

3. In a sweet shop, I spend £3.20 on three cans of soft drink and four bars of chocolate. Thenext day, I buy a can of soft drink and four bars of chocolate for £2. How much does each itemcost?



Example 8

In my pocket I have only 5-cent and 10-cent coins. How many of each type of coin do I have ifI have 24 coins altogether and their total value is $1.55?

Let x be the number of 5-cent coins and y be the number of 10-cent coins.

∴ x+ y = 24 .....(1) (the total number of coins)5x+ 10y = 155 .....(2) (the total value of coins)

Multiplying (1) by −5 gives

−5x − 5y = −1205x + 10y = 155

5y = 35

∴ y = 7

Substituting in (1) gives

x+ 7 = 24

∴ x = 17

Check: 17 + 7 = 24 X 5 · 17 + 10 · 7 = 155 X

Thus I have 17 five cent coins and 7 ten cent coins.

Exercises - Set F

1. I collect only 50-cent and $1 coins. My collection consists of 43 coins and their total value is$35. How many of each type of coin do I have?

2. Amy and Michelle have $29.40 between them and Amy’s money is three quarters of Michelle’sHow much money does each have?

3. margarine is sold in either 250 g or 400 g packs. A supermarket manager ordered 19.6 kg ofmargarine and received 58 packs. How many of each type did the manager receive?

4. Given that the triangle alongside is equilateral, find a and b.

(4a− b) cm

(a+ 4) cm(b+ 2) cm

5. A rectangle has perimeter 32 cm. If 3 cm is taken from the length and added to the width,the rectangle becomes a square. Find the dimensions of the original rectangle.

6. Twice the larger of two numbers is three more than five times the smaller, and the sum offour times the larger and three times the smaller is 71. What are the numbers?

7. 230 students and 29 staff are going on a school trip. They travel by large and small coaches.The large coaches seat 55 and the small coaches seat 39. If there are no spare seats and fivecoaches are to make the journey, how many of each coach are used?

8. Three times the width of a certain rectangle exceeds twice its length by three inches, andfour times its length is twelve more than its perimeter. Find the dimensions of the rectangle.



5. Harder problems

Example 9

A boat travels 24 km upstream in 4 hours. The return trip downstream takes only 3 hours.Given that the speed of the current is constant throughout the entire trip, what was:a) the speed of the current b) the speed of the boat in still water?

Let the speed of the boat in still water be x km per hour, and the speed of the current be y kmper hour.

speed upstream=24 km

4h= 6 km/h speed downstream=

24 km

3h= 8 km/h

boat speed − current speed = actual speed (upstream) ∴ x− y = 6 ..... (1) and

boat speed + current speed = actual speed (downstream) ∴ x+ y = 8 ..... (2)

adding (1) and (2): 2x = 14∴ x = 7

and substituting in (1): 7− y = 6∴ y = 1

Check: 7− 1 = 6 X 7 + 1 = 8 X

a) Current speed is 1 km/h b) Boat speed in still water is 7 km/h.

Exercises - Set G

1. A motor boat travels 12 km/h upstream against the current and 18 km/h downstream withthe current. Find the speed of the current and the speed of the motor boat in still water.

2. A jet plane made a 4000 km trip with the wind, in 4 hours, but required 5 hours to makethe return trip. Given that the speed of the wind was constant throughout the entire trip, whatwas the speed of the wind and what was the average speed of the plane in still air?

3. A man on foot covers the 25 km between two towns in 3 3

4hours. He walks at 4 km/h for

the first part of the journey and runs at 12 km/h for the remaining part.a) How far did he run? b) For how long was he running?

4. Explain why any two digit number can be written in the form 10a + b. Hence, solve thefollowing problem:A number consists of two digits which add up to 9. When the digits are reversed, the originalnumber is decreased by 45. What was the original number?

5. The difference between the two diagonals of a rhombus is 6 cm and its area is 56 cm2. Findthe dimensions of the diagonals.

6. The sum of the two digits of a number is 7. It you invert the order of the digits, the numberyou get is two more than twice the original number. What is the original number?



Review exercises

1. Solve the following simultaneous equations:

a)y = 2x− 53x− 2y = 11

b)3x+ 5y = 14x− 3y = 11

c)3x− 2y = 16y = 2x− 10

d)3x− 5y = 114x+ 3y = 5

2. Flour is sold in 5 kg and 2 kg packets. The 5 kg packets cost $2.75 and the 2 kg packetscost $1.25 each. If I bought 67 kg of flour and the total cost was $38.50, how many of eachkind of packet did I buy?

3. For the given rectangle, find its perimeter. Your answer must not contain x and y.

3x− y + 16

2x+ 3y + 5

2x− y + 6x+ y + 3

4. A bus company uses two different sized buses. If the company uses 7 small buses and 5 largebuses to transport 331 people, but needs 4 small buses and 9 large buses to carry 398 people,determine the number of people each bus can carry.

5. For this isosceles triangle the perimeter is 29 cm. Find the length of the equal sides. Youranswer must not contain x and y.

3x− 2y + 14

x+ y + 1

2x− y + 4

6. Orange juice can be purchased in 2 l cartons or in 600 ml bottles. The 2 l cartons cost $1.50each and the 600 ml bottles cost $0.60 each. A consumer purchases 73 l of orange juice andhis total cost was $57. How many of each container did the consumer buy?

7. Solve simultaneously:

a)

3x+ 3y

2−

3(x− y)

4= 6

x+ y − (2x− y) = −3

b)

x+ 1

3−

y − 1

2= 1

7x− 4(x+ y) = 4

c)

x− y = 1

2

5x+

3

4y = 5

d)

x

3+

y

5= 7

x

3−

y

4= −1

e)

x

2+

y

3=

−1

6x

2−

y

5=

1

10

f)

y

x−

1

2= 0

x

3+

y

2=

4

3

g)

−2(x+ 1) + 3(2y − 4) = 16

x

3−

y

2= −3

✁ ✃ ✁ ✃ ✁ ✃ ✁ ✃ ✁ ✃ ✁ ✃ ✁ ✃


Quadratic equations

1. Introduction

Equations of the form ax + b = 0 where a 6= 0 are called linear equations and have only one

solution.

For example, 3x− 2 = 0 is the linear equation with a = 3 and b = −2. It has the solution x =2

3.

Equations of the form ax2 + bx + c = 0 where a 6= 0 are called quadratic equations. Theymay have two, one or zero solutions.

Here are some simple quadratic equations which clearly show the truth of this statement:

Equation ax2 + bx+ c = 0 form a b c Solutions

x2 − 4 = 0 x2 + 0x− 4 = 0 1 0 −4 x = 2 or x = −2 two(x− 2)2 = 0 x2 + 4x+ 4 = 0 1 −4 4 x = 2 onex2 + 4 = 0 x2 + 0x+ 4 = 0 1 0 4 none as x2 is always ≥ 0 zero

Now consider the example x2 + 3x− 10 = 0.

If x = 2, then x2 + 3x− 10 = 22 + 3 · 2− 10 = 0, and

if x = −5, then x2 + 3x− 10 = (−5)2 + 3 · (−5)− 10 = 0

x = 2 and x = −5 both satisfy the equation x2 + 3x − 10 = 0, so we say that they are bothsolutions.

In this lesson we will learn methods for solving quadratic equations without using trial and error, andapply them to practical problems.

2. Quadratic equations of the form x2= k

Consider the equation x2 = 7.

Now√7 ·

√7 = 7, so x =

√7 is one solution,

and (−√7) · (−

√7) = 7, so x = −

√7 is also a solution.

Thus, if x2 = 7, then x = ±√7. (±

√7 is read as “plus or minus the square root of 7”).

Solution of x2= k

If x2 = k then

x = ±√k if k > 0

x = 0 if k = 0there are no real solutions if k < 0

1

2 Quadratic equations - 3 o ESO

Example 1

Solve for x: a) 2x2 + 1 = 15 b) 2− 3x2 = 8

a) 2x2 + 1 = 15

∴ 2x2 = 14 (take 1 from both sides)

∴ x2 = 7 (divide both sides by 2)

∴ x = ±√7

b) 2− 3x2 = 8

∴ −3x2 = 6 (take 2 from both sides)

∴ x2 = −2 (divide both sides by − 3)

which has no solutions as x2 cannot be < 0.

Example 2

Solve for x: a) (x− 3)2 = 16 b) (x+ 2)2 = 11

a) (x− 3)2 = 16

∴ x− 3 = ±√16

∴ x− 3 = ±4

∴ x = 3± 4

∴ x = 7 or − 1

b) (x + 2)2 = 11

∴ x+ 2 = ±√11

∴ x = −2±√11

For equations of the form (x ± a)2 = k we do not expand the LHS.

Exercises - Set A

1. Solve for x:

a) x2 = 100 b) 5x2 = 20 c) 6x2 = 54 d) 5x2 = −45

e) 7x2 = 0 f) 3x2 − 2 = 25 g) 4− 2x2 = 12 h) 4x2 + 2 = 10

2. Solve for x:

a) (x− 1)2 = 9 b) (x+ 4)2 = 16 c) (x+ 2)2 = −1 d) (x− 4)2 = 5

e) (x+ 2)2 = 0 f) (2x− 5)2 = 0 g) (3x+ 2)2 = 4 h) 1

3(2x+ 3)2 = 2

3. The Null Factor law

For quadratic equations which are not of the form x2 = k, we need other methods of solution. Onemethod is to factorise the quadratic and then apply the Null factor law.


Quadratic equations - 3 o ESO 3

The Null Factor law states that:

When the product of two (or more) numbers is zero, then at least one of them must be zero.So, if ab = 0 then a = 0 or b = 0.

This is specially useful to solve quadratic equations of the form ax2 + bx = 0.

Example 3

Solve for x: a) 3x2 − 15x = 0 b) (x − 4)(3x+ 7) = 0

a) 3x2 − 15x = 0

∴ 3x(x− 5) = 0

∴ 3x = 0 or x− 5 = 0

∴ x = 0 or x = 5

b) (x − 4)(3x+ 7) = 0

∴ x− 4 = 0 or 3x+ 7 = 0

∴ x = 4 or 3x = −7

∴ x = 4 or x = − 7

3

Exercises - Set B

1. Solve for the unknown using the Null Factor law:

a) 3x = 0 b) −7y = 0 c) ab = 0 d) 2xy = 0

e) abc = 0 f) a2 = 0 g) pqrs = 0 h) a2b = 0

2. Solve for x using the Null Factor law:

a) x(x− 5) = 0 b) 2x(x+ 3) = 0 c) (x + 1)(x− 3) = 0

d) 3x(7− x) = 0 e) −2x(x+ 1) = 0 f) 4(x+ 6)(2x− 3) = 0

g) (2x+ 1)(2x− 1) = 0 h) 11(x+ 2)(x− 7) = 0 i) −6(x− 5)(3x+ 2) = 0

j) x2 = 0 k) 4(5− x)2 = 0 l) −3(3x− 1)2 = 0

Remember that in order to use the Null Factor law we must have one side of the equation equal to

zero.

Example 4

Solve for x: x2 = 3x

x2 = 3x

∴ x2 − 3x = 0 rearranging so RHS = 0

∴ x(x− 3) = 0 factorising the LHS

∴ x = 0 or x− 3 = 0 Null factor law∴ x = 0 or x = 3



Illegal cancelling

Let us reconsider the equation x2 = 3x from the previous example. We notice that there is a commonfactor of x on both sides.

If we cancel x from both sides, we will havex2

x=

3x

xand thus x = 3.

Consequently, we will lose the solution x = 0.

From this example we conclude that:

We must never cancel a variable that is a common factor from both sides of an equation unlesswe know that the factor cannot be zero.

Exercises - Set C

1. Solve for x:

a) x2 − 7x = 0 b) x2 − 5x = 0 c) x2 = 8x d) x2 = 4x

e) 3x2 + 6x = 0 f) 2x2 − 5x = 0 g) 4x2 = 5x h) 3x2 = 9x

4. Completing the square

A quadratic equation of the type ax2 + bx + c = 0 may not be so easy to factorise, then to solvethis type of equations we need a different technique.Let us consider for example the equation x2 + 4x− 7 = 0.

We saw in Example 2 that if (x+ 2)2 = 11, then x+ 2 = ±√11 and therefore x = −2±

√11.

So, we can solve x2+4x−7 = 0 if we can rearrange it so there is a perfect square on the left hand side:

x2 + 4x− 7 = 0

∴ x2 + 4x = 7

∴ x2 + 4x+ 4 = 7 + 4 (adding 4 to both sides to complete a perfect square on the LHS)

∴ (x+ 2)2 = 11

∴ x+ 2 = ±√11

∴ x = −2±√11

Hence the solutions to x2 + 4x− 7 = 0 are x = −2±√11.

The process of creating a perfect square on the left hand side is called completing the square.

From our previous study of perfect squares we observe that:

(x + 3)2 = x2 + 2 · 3 · x+ 32

(x − 5)2 = x2 − 2 · 5 · x+ 52

(x+ p)2 = x2 + 2 · p · x+ p2

So, the constant term is “the square of half the coefficient of x”.



Example 5

To create a perfect square on the LHS, what must be added to both sides of the equation:

a) x2 + 8x = −5 b) x2 − 6x = 13?

What does the equation become in each case?

a) In x2 + 8x = −5, half the coefficient of x is 8

2= 4. So, we add 42 to both sides and the

equation becomes

x2 + 8x+ 42 = −5 + 42

∴ (x+ 4)2 = −5 + 16

∴ (x+ 4)2 = 11

b) In x2 − 6x = 13, half the coefficient of x is −6

2= 3. So, we add (−3)2 = 32 to both sides

and the equation becomes

x2 − 6x+ 32 = 13 + 32

∴ (x− 3)2 = 13 + 9

∴ (x− 3)2 = 22

Example 6

Solve for xby completing the square, leaving answers in simplest radical form:

a) x2 + 2x− 2 = 0 b) x2 − 4x+ 6 = 0

a) x2 + 2x− 2 = 0

∴ x2 + 2x = 2 move constant term to RHS

∴ x2 + 2x+ 12 = 2 + 12 add 12 to both sides

∴ (x+ 1)2 = 3 factorise LHS, simplify RHS

∴ x+ 1 = ±√3

∴ x = −1±√3

b) x2 − 4x+ 6 = 0

∴ x2 − 4x = −6 move constant term to RHS

∴ x2 − 4x+ 22 = −6 + 22 add 22 to both sides

∴ (x− 2)2 = −2 factorise LHS, simplify RHS

which is impossible as no perfect square can be negative. Hence, no real solutions exist.

Exercises - Set D

1. For each of the following equations find what must be added to both sides of the equationto create a perfect square on the LHS, and write each equation in the form (x+ p)2 = k:

a) x2 + 2x = 5 b) x2 − 2x = −7 c) x2 + 6x = 2

d) x2 − 6x = −3 e) x2 + 10x = 1 f) x2 − 8x = 5

g) x2 + 12x = 13 h) x2 + 5x = −2 i) x2 − 7x = 4



Exercises - Set E

1. If possible, solve for x by completing the square, leaving answers in simplest radical form:

a) x2 − 4x+ 1 = 0 b) x2 − 2x− 2 = 0 c) x2 − 4x− 3 = 0

d) x2 + 2x− 1 = 0 e) x2 + 2x+ 4 = 0 f) x2 + 4x+ 1 = 0

g) x2 + 6x+ 3 = 0 h) x2 − 6x+ 11 = 0 i) x2 + 8x+ 14 = 0

j) x2 + 3x+ 2 = 0 k) x2 = 4x+ 8 l) x2 − 5x+ 6 = 0

m) x2 + x− 1 = 0 n) x2 + 3x− 1 = 0 o) x2 + 5x− 2 = 0

5. The quadratic formula

Many quadratic equations cannot be solved by factorisation, and completing the square is rathertedious. Consequently, the quadratic formula has been developed.

If ax2 + bx+ c = 0 where a 6= 0, then x =−b±

√b2 − 4ac

2a.

Proof: If ax2 + bx+ c = 0

then x2 +b

ax+

c

a= 0 dividing each term by a, as a 6= 0

∴ x2 +b

ax = − c

a

∴ x2 +b

ax+

(

b

2a

)2

= − c

a+

(

b

2a

)2

completing the square on LHS

∴

(

x+b

2a

)2

= − c · 4aa · 4a +

b2

4a2

∴

(

x+b

2a

)2

=b2 − 4ac

4a2

∴ x+b

2a= ±

√

b2 − 4ac

4a2

∴ x = − b

2a±

√b2 − 4ac

2a

∴ x =−b±

√b2 − 4ac

2a

Example 7

Solve for x: x2 − 2x− 2 = 0

x2 − 2x− 2 = 0 has a = 1, b = 2, c = −2, then

x =−(−2)±

√

(−2)2 − 4(1)(−2)

2(1)=

2±√4 + 8

2=

2±√12

2=

2± 2√3

2= 1± 3



Example 8

Solve for x: 2x2 + 3x− 4 = 0

2x2 + 3x− 4 = 0 has a = 2, b = 3, c = −4, then

x =−3±

√

32 − 4(2)(−4)

2(2)=

−3±√9 + 32

4=

−3±√41

4

Exercises - Set F

1. Use the quadratic formula to solve for x:

a) x2 + 9x+ 14 = 0 b) x2 + 5x+ 6 = 0 c) x2 − 3x+ 2 = 0

d) x2 − 5x+ 6 = 0 e) x2 + 2x = 2 f) x2 + 2 = 6x

g) x2 = 4x+ 1 h) x2 + 1 = 3x i) x2 + 8x+ 5 = 0

j) 2x2 + 7x+ 5 = 0 k) 3x2 + 13x+ 4 = 0 l) 5x2 = 13x+ 6

m) 9x+ 36 = x2 n) 2x2 = 2x+ 1 o) 9x2 = 6x+ 1

p) 25x2 + 1 = 20x q) 2x2 + 6x+ 1 = 0 r) 3x2 + 2x− 2 = 0

2. Solve for x by first expanding brackets:

a) x(x+ 5) + 2(x+ 6) = 0 b) x(1 + x) + x = 3 c) (x− 1)(x+ 9) = 8x

d) 3x(x+ 2)− 5(x− 3) = 17 e) 4x(x+ 1) = −1 f) 2x(x− 6) = x− 20

g) (x+ 2)(x− 1) = 5 h) (x+ 1)2 = 3− x2

3. Solve for x by first eliminating the algebraic fractions:

a)x

3=

2

xb)

x− 1

4=

3

xc)

x− 1

x=

x+ 11

5

d)2x

3x+ 1=

1

x+ 2e)

2x+ 1

x= 3x f)

x+ 2

x− 1=

x

2

6. The discriminant, ∆

Consider x2 + 2x+ 5 = 0. Using the quadratic formula, the solutions are:

x =−2±

√

4− 4(1)(5)

2(1)=

−2±√−16

2

However, in the real number system,√−16 does not exist. We therefore say that x2 +2x+5 = 0

has no real solutions.

Consider now x2 + 4x+ 4 = 0. Using the quadratic formula, the solutions are:

x =−4±

√

16− 4(1)(4)

2(1)=

−4±√0

2=

−4

2= −2

As we get√0 = 0, there is only one solution.



In the quadratic formula, the quantity b2−4ac under the square root sign is called the discriminant.The symbol delta or ∆ is used to represent the discriminant, so ∆ = b2 − 4ac.

Notice that:

• if ∆ = 0, x =−b

2ais the only solution and is known as a repeated root.

• if ∆ > 0,√∆ is a real number and so there are two distinct real roots

−b+√∆

2aand

−b−√∆

2a

• if ∆ < 0,√∆ does not exist and so we have no real solution.

Example 9

Use the discriminant to determine the nature of the roots of:

a) 2x2 − 3x+ 4 = 0 b) 4x2 − 4x− 1 = 0

a) ∆ = b2 − 4ac = (−3)2 − 4 · 2 · 4 = −23 < 0, so there are no real roots.

b) ∆ = b2 − 4ac = (−4)2 − 4 · 4 · (−1) = 32 > 0, so there are two distinct real roots.

Exercises - Set G

1. Find the discriminant of:

a) x2 − 2x− 7 = 0 b) 2x2 − 3x+ 6 = 0 c) x2 − 11 = 0

d) 2x2 − 6x− 4 = 0 e) 3x2 + 7x− 1 = 0 f) 4x2 − 7x+ 11 = 0

2. Using the discriminant only, state the nature of the solutions of:

a) x2 + 7x− 2 = 0 b) x2 + 4√2x+ 8 = 0 c) 2x2 + 3x− 1 = 0

d) 6x2 + 5x− 4 = 0 e) x2 + x+ 6 = 0 f) 9x2 + 6x+ 1 = 0

3. Show that the following quadratic equations have no real solutions:

a) x2 − 3x+ 12 = 0 b) x2 + 2x+ 4 = 0 c) −2x2 + x− 1 = 0

4. Solve for x, where possible:

a) x2 − 25 = 0 b) x2 + 25 = 0 c) x2 − 7 = 0

d) x2 + 7 = 0 e) 4x2 − 9 = 0 f) 4x2 + 9 = 0

g) x2 − 4x+ 5 = 0 h) x2 − 4x− 5 = 0 i) x2 − 10x+ 29 = 0

j) x2 + 6x+ 25 = 0 k) 2x2 − 6x− 5 = 0 l) 2x2 + x− 2 = 0

7. Problem solving

The problems in this section can all be converted to algebraic form as quadratic equations, thatcan be solved as in the previous sections.



Problem solving method

Step 1: Carefully read the question until you understand the problem. A rough sketch may beuseful.

Step 2: Decide on the unknown quantity and label it x, say.

Step 3: Use the information given to find an equation which contains x.

Step 4: Solve the equation.

Step 5: Check that any solutions satisfy the equation and are realistic to the problem.

Step 6: Write your answer to the question in sentence form.

Example 10

The sum of a number and its square is 42. Find the number.

Let the number be x. Therefore its square is x2.

x+ x2 = 42

∴ x2 + x− 42 = 0

∴ x =−1±

√

12 − 4 · 1 · (−42)

2

∴ x = −7 or x = 6

So, the number is −7 or 6.

Check: If x = −7, −7 + (−7)2 = 42 X, and if x = 6, 6 + 62 = 42 X

Example 11

A rectangle has length 5 cm greater than its width. If it has an area of 84 cm2, find thedimensions of the rectangle.

If x cm is the width, then (x+ 5) cm is the length.

Now area = 84 cm2

∴ x(x+ 5) = 84

∴ x2 + 5x = 84

∴ x2 + 5x− 84 = 0

∴ x =−5±

√

52 − 4 · 1 · (−84)

2

∴ x = −12 or x = 7

But x > 0 as lengths are positive quantities, so x = 7. Hence the rectangle is 7 cm by 12 cm.



Exercises - Set H

1. The sum of a number and its square is 110. Find the number.

2. The product of a number and the number increased by 4 is 117. Find the two possible valuesof the number.

3. When 24 is subtracted from the square of a number, the result is five times the originalnumber. Find the number.

4. The sum of two numbers is 6 and the sum of their squares is 28. Find the exact values ofthese numbers.

5. Two numbers differ by 7 and the sum of their squares is 29. Find the numbers.

6. A rectangle has length 4 cm greater than its width. Find its width given that its area is 96cm2.

7. A triangle has base 4 m more than its altitude. If its area is 70 m2, find its altitude.

8. A rectangular enclosure is made from 60 m of fencing. The area enclosed is 216 m2. Findthe dimensions of the enclosure.

9. A rectangular garden bed was built against an existing brick wall. 24 m of edging was usedto enclosure 60 m2. Find the dimensions of the garden bed to the nearest cm.

10. ABCD is a rectangle in which AB = 21 cm. The square AXYD is removed and theremaining rectangle has area 80 cm2. Find the length of BC.

A X B

D Y C

11. Sarah has a brother who is 5 years younger than herself and another brother who is 8 yearsyounger than herself. She observes that the product of her brothers’ ages is equal to the age ofher 40 year old father. How old is Sarah?

12. The numerator of a fraction is 3 less than the denominator. If the numerator is increasedby 6 and the denominator is increased by 5, the fraction is doubled in value. Find the originalfraction.

13. 182 sweets are equally divided among a certain number of children at a party. If the numberof sweets each child receives is one more than the number of children, find the number ofchildren at the party.

14. The sum of a number and its reciprocal is 2 1

12. Find the number.

15. The sum of a number and twice its reciprocal is 3 2

3. Find the number.

16. In a 42 km marathon race a runner took (n−4) hours to run the race, running at a constantspeed of (n + 7) km/h. Find the time the runner took to run the race and the speed of therunner.

✁ ✃ ✁ ✃ ✁ ✃ ✁ ✃ ✁ ✃ ✁ ✃ ✁ ✃


Geometry

• Pythagoras’ theorem

• Thales’ theorem = Intercept theorem. Similar triangles

• Similarity

• Transformation geometry

• Geometric solids

• The Earth. Meridians and parallels. Word problems

Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona Pythagoras’ theorem 1

Pythagoras’ theorem

A right angled triangle is a triangle which has a right angle as

one of its angles.

The side opposite the right angle is called the hypotenuse and is

the longest side of the triangle.

The other two sides are called the legs of the triangle.

Around 500 BC, the Greek mathematician Pythagoras discovered a rule which connects the

lengths of the sides of all right angled triangles

Pythagora’s

theorem:

In a right angled triangle with

hypotenuse � and legs � and�,

�� .

EXAMPLE 1

A gardener marks out a new lawn that is supposed to be a rectangle with sides of lengths

8 m and 12 m. He checks that he has marked out a rectangle by measuring the diagonal.

How long should the diagonal be, correct to 1 decimal place?

�� 8� � 12�

� � √64 � 144 � √208 � 14.4 m.

EXAMPLE 2

A rope, of length 30 m, is tied to the top of a flagpole. The height of the flagpole is 15 m.

How far will the end of the rope be from the base of the flagpole, if it is pulled tight,

correct to 1 decimal place?

30� � �� 8�

900 � �� 64

836 � ��

� � √836 � 28.9 m.

8

12

d

15 30

x

Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona Pythagoras’ theorem 2

EXERCISES

1. Decide whether each of the following triangles contains a right angle:

2. Determine the length of the diagonal of a square with sides of length 20 cm.

3. An isosceles triangle has two sides of length 10 cm and one side of length 4 cm. Calculate the

perpendicular height of the triangle.

4. Calculate the perimeter of the following triangle, giving your answer correct to 2 decimal places:

5. The diagram shows the side view of a shed. What is the length of the sloping roof of the shed,

correct to 2 decimal places?

6. A boat sails due east for 20 km and it then sails north for 5 km. How far in km is the boat from

its starting position, correct to 1 decimal place?

7. A ladder leans against a vertical wall. The length of the ladder is 6 m. The bottom of the ladder

is 2 m from the base of the wall. How high is the top of the ladder above the ground, correct to

2 decimal places?

8. An equilateral triangle has area 16√3 cm2. Find the length of its sides.

9. An extension ladder rest 4 m up a wall. If the ladder is extended a further 0.8 m without moving

the foot of the ladder, then it will now rest 1 m further up the wall. How long is the extended

ladder?

Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona Intercept theorem 1

Intercept theorem (Thales´ theorem)

Related concepts:

Similarity and similar triangles

Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona Intercept theorem 2

EXAMPLE 1

In the diagram given, find �:

Using Thales´theorem (Intercept theorem):

AC and AD are two intersecting lines. BE and

CD are two parallel lines that intercept AC and

AD, and so

|��|

|��|=|��|

|��|

8

4=�

6

� =6 ∙ 8

4= 12��

Using similarity:

∆s ACD and ABE are similar, and so

|��|

|��|=|��|

|��|

8 + 4

8=� + 6

�

��8 + 4� = 8�� + 6�

8� + 4� = 8� + 48

4� = 48

� = 12��

EXERCISES

1. In the following, find the values of the unknows.

a)

b)

Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona Intercept theorem 3 c)

d)

e)

f)

2. In the figure, AF//BE//CD. If AB=15 cm, BC=5cm and FD=40 cm, find the length of FE.

3. Find the length of AC.

Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona Sim ilarity 1

Similarity

Two figures are similar if one is an enlargement of the other (regardless of orientation)

If two figures are similar then their corresponding sides are in proportion. This means that the

lengths of sides will be increased (or decreased) by the same ratio from one figure to the next.

This ratio is called the enlargement factor. We often denote it by k.

Consider the enlargement below for which the enlargement factor is 1.5

Since k= 1.5, notice that �´�´�� = �´�´

�� = �´�´�� = �´�´

�� = �´�´�� = �.

If two figures are similar then:

• The figures are equiangular, and

• The corresponding sides are in proportion

Similar triangles

If two triangles are equiangular then they are similar.

Similar triangles have corresponding sides in the same ratio

To test if two triangles are similar, we need to show that:

• At least two pairs of angles are equal in size or

• Their sides are in proportion


EXAMPLE 1

In the diagram given, establish that a that a pair of triangles is similar and find :

We label the vertices of the figure so

that we can easily refer to them

∆s ABE and ACD are similar as

� = �� = �� {corresponding angles}

Corresponding sides must be in the same ratio, and so

�� = ��

��

∴ 66 + 3 =

7

∴ 7 = 69

∴ = 69 ∙ 7

∴ = 4.67

Similar polygons

Two figures are similar if they have the same shape. The figures may not necessarily be the

samae size. Two polygons are similar if all corresponding angles are congruent and the

measures of all corresponding sides form the same ratio (are in proportion).

The symbol for similarity is ~

Polygon ABCDE~ polygon KLMNO

Corresponding angles are congruent (identical)

Corresponding sides are in proportion "#�� = #$

�� =$%�� = %&

�' =&"'� = (


EXERCISES 1

1 For the following figures, establish taht a pair of triangles is similar, and hence find or).

2,. In each figure, find the value of x to the nearest tenth.

EXERCISES 2

1. A ramp is buit to enable wheel-chair access to a building that is 24 cm above ground level.

The ramp has a constant slope of 2 in 15, which means that for every 15 cm horizontally

rises 2 cm. Calculate the length of the base of the ramp


2.

3.

4. At the same time as the shadow cast by a vertical 30 cm long ruler is 45 cm

long. Peter`s shadow is 268 cm long.

a) Draw a fully labeled sketch of the situation

b) Find Peter`s height

5.

6. María placed a mirror on the ground and stood so that she could see the top of

the tree. What is the height of the tree?


Areas and volumes of similar figures

RELATED AREAS

Consider the triangle illustrated with

base * and altitude+ cm. If the base

and altitude are multiplied by the

same positive constant,, a figure is

obtained which is similar.

Notice that:

�-.+(()�-.+(�) =

�( (*,)(+,)

�(*+

= ,(

This suggests that:

If the corresponding sides of similar figures are in the ratio ,, then

Area of image = ,( × +-.+232*4.56.

RELATED VOLUMES

If the sides of a rectangular prism are all

multiplied by k then

7289:.(()7289:.(�) =

,;+*5+*5 = ,;

This suggests that:

If the corresponding sides of similar solids are in the ratio ,, then

Volume of image = ,; × +-.+232*4.56.

EXAMPLE 2

Find the unknown area. (The triangles are similar).

�<. ( = =>?

(

� = 6.2 × 8125 = 20,088FG�


EXAMPLE 3

Triangle DEC has area 4.2 cm2.

a) Find the area of triangle ABC

b) Find the area of quadrilateral ABED

The ∆s DEC and BAC are equiangular and so are similar

a) HIJH∆KLMHIJH∆NOM = PQRS

� = 1.96

�TU�∆��4.2FG� = 1.96

�TU�∆�� = 1.96 × 4.2

= 8.232FG�

b) �TU�VW�� = (8.232 − 4.2) = 4,032 cm2

.

EXERCISES 3

1 For each pair of similar figures, find the unknown area.

a)

b)

c)

d)

Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona Sim ilarity 7 2 In the given figure, the area of ∆BCD is 9.3 cm

2. Find the area of

a) ∆ACE

b) quadrilateral ABDE

3 Find

a) The value of x

b) The area of ∆PQT given quadrilateral QRST has area 22 m2

4 What will happen to the volume of

a) a sphere if the radius is doubled

b) a sphere if the radius is increased by 20%

c) a cylinder if the radius and height are halved

d) a cylinder if the radius and height are increased by 50%

RATIO and SCALE

The ratio of the length in a drawing (or model) to the length of the real thing.

When we make a scale copy of an object, the original and the copy have the same proportions.

We can consider the copy and the real thing similar figures.

YZ[6+\5.2\6].-.+86]Z\^2-2*4.56(9\Z6[)YZ[6+\5.2\6].:2Y.82-:+_([+:.9\Z6[) = [5+8.3+562-

Or using proportions

�[5+8.3+562- = YZ[6+\5.2\:2Y.8(9\Z6[)

YZ[6+\5.2\-.+86]Z\^([+:.9\Z6[)

To describe how much an object has actually been scaled down (or up), we often use ratios. So

a scale of 1:12 means that for every 1 cm that a particular part of the model measures, the

corresponding part of the real thing measures 12 cm. This also means that the object is 12

times the length of the model.

For example:

SCALE 1 : 200 means

1 cm on the

model

200 cm on the real thing


Example: a scale of 1:10 means that in the drawing anything with the size of "1" would have

a size of "10" in the real world, so a measurement of 150 mm on the drawing would be 1500

mm on the real thing.

EXAMPLE 4

The distance between two towns measures 6 cm on a map. What is the true distance if the scale

is 1 : 50 000

the scale is 1 : 50 000, so 1 cm on the map is 50 000 cm in actual length

The True Distance = 6 cm x 50 000

= 300 000cm

We can simplify this by dividing the 300 000 by 100 which gives us 3000m

and by further dividing the 3000m by 1000 we can simplify this to 3km

Therefore the True Distance between the two towns is 3km.

Using proportions:

` = < × aaaa = ;aaaaa5:

` = ;aaaaa5:�aa5:/: = ;aaa:

` = ;aaa:�aaa:/,: = ;,:


EXAMPLE 5

The drawing of an Aircraft uses a scale of 1 : 900 If the Aircrafts wingspan is 30m what length on the

drawing actually represents this?

30 m = 30 x 100 cm = 3000 cm

1900 =

FG3000FG

= 3000FG900 = 3.333FG

= 3.333FG × 10GGFG = 33.33GG

Therefore the length on the drawing that represents the wingspan is 33.33 mm

Exercises

1.

2. The length of a football stadium is 105 m and its width is 68 m, and you want to represent it

in your notebook. So, you need to make a reduction!

a) Find a suitable scale factor (decide first what length is suitable for your drawing, and

from this decision you can get the scale factor)

b) Calculate the length and width of the football stadium in your drawing

c) Represent it in your notebook


3. Here is the floor plan of a country house.

In bedroom 1 you have the dimension of the width of that bedroom. From this:

a) Calculate the scale factor

b) Find out the dimensions (width and length) of bedrooms, bathrooms, kitchen and

closets. Show all your work and reasoning.

c) Find out the dimensions of the entry and the living-dining room. Explain clearly your

method to determine the surface area of the bay windows.

d) Calculate the whole floor surface area.

4.


5


6

Departamento de Matemáticas. Real Instituto de Jovellanos. J. F. Antona Transformation geometry 1

Transformation geometry

A change in the size, shape, orientation or position of an object is called a transformation.

Reflections, rotations, translations and enlargements are all examples of transformations. We can describe these transformations mathematically using transformation geometry.

Many trees, plants, flowers, animals and insects are symmetrical in some way. Such symmetry results from a reflection, so we can describe symmetry using transformations.

In transformation geometry figures are changed (or transformed) in size, shape, orientation or position according to certain rules.

The original figure is called the object and the new figure is called the image.

We will consider the following transformations:

• Translations where every point moves a fixed distance in a given direction

• Reflections or mirror images

• Rotations about the origin O through a given angle


1.- TRANSLATIONS

A translation moves an object from one place to another. Every point on the object moves the same distance in the same direction.

If �(�, �) is translated � units in the �-direction and� units in the �-direction to become �´(�´, �´), then �´ = � + �and�´ = � + �

We write Point �(�, �) moved under a translation with vector ��′�� (h, k) to get image

P’(x+h, y+k)

Where P’ is the image of the object P and

��′�� (h, k) is called the translation vector.

�� = � + ��′ = � + �� are called the transformation equations.

Translation vectors are written too, using small letters with an arrow on the top. Ex ��(h,k)

In some books, translations vectors are written as �ℎ�� .


EXERCISE 1

1.

2.

3.

4.

5.


EXAMPLE 1

Triangle OAB with vertices O(0,0), A(2,3) and B(-1,2) is translated 3 units in the x-direction and 2 units in the y-direction. So the translation vector is ��(3,2). Find the image vertices and illustrate the object and image.

O’(3,2) because O +�� = O’

Using the transformation equations: �� = � + ��′ = � + ��

�� + � = �� + � = ��

A’(5,5) because A +�� =A �� + � = �� + � = ��

B’(2,4) because B +�� =B’ �� + � = �� + � = �

EXAMPLE 2

Find the image equation when �� = ! is translated ��(-1,2). Check your result by graphing.

The transformation equations are: �� = � + ��′ = � + ��

So, �� = � � �"#$�� = � + �

∴ � = �� + �"#$� = ��

So, we replace � by (� + �) and �by (� � �), And �� = ! becomes

�(� + �) � �(� � �) = !

∴ �� + � � �� + ! = !

∴ �� = ��

The image line is parallel to the object line


EXERCISE 2

1. Find the object or image coordinates, or translation vector components in the following:

Object

coordinates

Translation

vector

components

Image

coordinates

(5, 4) (6, 3)

(-3, -2) (-2, 5)

(2,-1) (6,-4)

(2,-4) (3, -1)

(-1,5) (0,-2)

(3, 7) (-1,5)

(3,8) (2,1)

(-3,-6) (-9,-11)

(5, 0) (-4, -2)

2. Find the image equations of the following, and if possible give your answers in the form & = '((). a) 3x+2y=8 under ��(-1,3).

b) x=4 under ��(2,1).

c) 2x-y=6 under ��(-3,0).

d) y=5 under ��(2,-5).

e) y=x2 under ��(0,3).

f) Y=-2x2 under ��(3,2).

SUCCESSIVE TRANSLATIONS.

Successive transformations are defined as more than one transformation being applied to an object. For example, a translation is applied to A and then a second translation is applied to A′ creating A′′ .

The resulting translation vector is the addition of each individual translation vector.


Successive translations on a triangle:

The components of the total translation guide vector can be found adding “like components” of the individual translation vectors.

In the above drawing, we have a triangle whose coordinates correspond to points A(1,2), B(2,4) and C(5,1).

We will perform successive translations using vectors )��(�, ) and ��(*, ��) as guide vectors.

We calculate the sum of “like components” of these vectors:

)�� + �� = +�� The components of +�� are (� + *, + (��)) = (��, �) The guide vector is represented in red in the above drawing: )�� + �� = +�� = (��, �) To calculate the equivalent counterpart points of the triangle vertices, we add the coordinates of each point with the guide vector:

Coordinates of A’’ = (�, �) + (��, �) = (� + ��, � + �) = (��, ,) Coordinates of B’’ = (�, ) + (��, �) = (� + ��, + �) = (��, *) Coordinates of C’’ = (,, �) + (��, �) = (, + ��, � + �) = (�,, )


EXERCISE 3

1. a) Find the resulting vector components if point P is translated under -��(2,3) followed by

translation ��(3,5) to get P’’(10,12)

b) Find P coordinates.

2. The resulting vector of two consecutive translations is 0�� = (7, 10). If one translation has used

vector -��(7,10), what are the components for the other translation guide vector?

3. Triangle ABC has as apexes A(3,5), B(5,7) and C(5,2). Find the coordinates of the apexes of the

translated triangle under two successive translations with guide vectors -��(6, 2) and ��(7, �2).

2.- REFLECTIONS

When P(x,y) is reflected in the mirror line to become P’(x’,y’), the mirror line perpendicularly bisects [PP’].

Thus, for every point on an object, the mirror line perpendicularly bisects the line segment joining the point with its image.

• Symmetry around an axis is Axial symmetry (“simetria axial” in Spanish). The mirror line is the axis in the above drawing.

• Symmetry around a point is Point symmetry (“simetria central” in Spanish) . Point Symmetry is when every part has a matching part that is ...

1.- the same distance from the central point ... 2.- but in the opposite direction.

It looks the same when viewed from opposite directions, such as left vs. right, or if turned upside down. It is sometimes called Origin Symmetry


Some examples of point symmetry:

EXERCISE 4

1.

2.

3.

4.

Departamento de Matemáticas. Real Instituto de Jovellanos. J. F. Antona Transformation geometry 9 5. Do a drawing of a triangle with coordinates of the apexes A(1,1), B(4,5) and C(6,4). Find the new

coordinates under:

a) reflection in the x-axis b) reflection in the y-axis c) Point reflection trough O (origin of coordinates)

6. Do a drawing of a triangle with coordinates of the apexes A(-5,5), B(-3,0) and C(-9,1). Find the new coordinates under:

a) reflection in the x-axis b) reflection in the y-axis c) Point reflection trough O (origin of coordinates)

We will concentrate on the following reflections (regarding axial symmetry):

M� the reflection in the x-axis

M� the reflection in the y-axis

M � = � the reflection in the line � = � M � = �� the reflection in the line � = ��

EXAMPLE 3

Find the image of the point (3,19 in:

a) M( b) M& c) M & = ( d) M & = �(

a) (3,1) 789::; (3,-1)

b) (3,1) 7< 9:; (-3,1)

c) (3,1) 7<=89::::; (1,3)

d) (3,1) 7<=�89:::::; (-1,-3)


The above diagram is useful for deducing the general equations of the four basic reflections.

For example, from (3,1) 7<=89::::; (1,3) we can see

That in general (x,y) 7<=89::::; (y,x).

∴ �´ = � and �´ = �

(�, �) 7�9::;(�, ��) (�, �) 7�9:;(��, �)(�, �) 7�=�9::::;(�, �)(�, �) =�=��9::::::;(��,��)

∴ �´ = � and �´ = ��

∴ �´ = �� and �´ = �

∴ �´ = � and �´ = �

∴ �´ = �� and �´ = ��

EXAMPLE 4

Find the image equation of �� = > reflected in the y-axis

The transformation equations are: �´ = ��"#$�´ = � ∴ � = ��´"#$� = �´

So, we replace � by (��) and leave � as is,

and �� = > becomes �(��)– �(�) = > ∴ �� = > ∴ �� + �� = �>

EXERCISE 5

1. Find by graphical means, the image of:

a) (4, �1) under Mx

b) (4, �1) under My

c) (4, �1) under My=x

d) (4, �1) under My=-x

e) (�1,�3) under Mx

f) (�1,�3) under My

g) (�1,�3) under My=x

h) (�1,�3) under My=-x

2. Find the image equation of:

a) & = 2( + 3 under Mx

b) & = (A under Mx

If you forget these or

want to check any of

them, choose a point

such as (3,1)

Departamento de Matemáticas. Real Instituto de Jovellanos. J. F. Antona Transformation geometry 11 c) & = B

C under My

d) & = 2C under My=x

e) 2( + 3& = 4 under My=-x

3. Find the image of:

f) D(2,3) under Mx followed by translation ��(�1, 2). g) D(4,�1) under My =-− x followed by translation ��(4,3). h) D(�1, 5) under My followed by Mx followed by translation ��(2,�4).

i) D(3,�2) under My=x followed by translation ��(3, 4). j) D(4, 3) under translation ��(1,�4) followed by Mx .

3.- ROTATIONS

When P(x,y) moves under a rotation about O through and angle of θ to a new position P’(x’,y’)

then OP = OP’ and PÔP’=θ where positive θ is measured anticlockwise.

O is the only point which does not move under the rotation.

Notation: Rθ means “a rotation about O through an angle of θ° ”.

We will concentrate on the following rotations:

(�, �) EF�9::;(��, �) (�, �) E�F�9:::;(�, ��)(�, �) E�>�9:::;(��,��)

∴ �´ = �� and �´ = �

∴ �´ = � and �´ = ��

∴ �´ = �� and �´ = ��


EXAMPLE 5

Find the image of the point (3,1) under: a)EF� b)EGF� c)E�>�

a)

(�, �) EF�9::;(��, �) anticlockwise

b)((((�, �) E�F�9:::;(�, ��) clockwise

c)(�, �) E�>�9:::;(��,��)

EXERCISE 6

1.

Departamento de Matemáticas. Real Instituto de Jovellanos. J. F. Antona Transformation geometry 13 2.

3.

4.

5.

6. Do a drawing of a triangle with coordinates of the apexes A(3,4), B(7,6) and C(9,1). Find the new

coordinates under:

a) Point reflection through O (origin of coordinates). b) 180º rotation c) Check the two results. What would you say?

EXERCISE 7

1. Find the image of the point:

a) (4, �1) under EHI, EGHIandEJKI

b) (�2, 3) under EHI, EGHIandEJKI

c) (2,6) under EHI, EGHIandEJKI

d) (�5,�7) under EHI, EGHIandEJKI

2. Find the image of:

a) D(2,3) under EHI followed by Mx

b) D(�2,5) under My =-− x followed by EGHI

c) D(�1, 5) under EJKI followed by Mx followed by translation ��(3,�2).

d) D(4,�2) under EHI followed by translation ��(2, 1). e) D(4, 3) under My =x followed by translation ��(1,�4) followed by EGHI.

GEOMETRIC SOLIDS POLYHEDROM: A solid formed by plane faces that are polygons.

POLIGON: A close plane figure bounded by straight lines.

In geometry, a Platonic solid is a convex polyhedron that is regular, in the sense of a regular

polygon. Specifically, the faces of a Platonic solid are congruent regular polygons, with the

same number of faces meeting at each vertex; thus, all its edges are congruent, as are its

vertices and angles.

There are precisely five Platonic solids (shown below).

��

��

The name of each figure is derived from its number of faces: respectively 4, 6, 8, 12, and 20.

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� ��

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The relationship between these

values is given by Euler's formula:

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�� !��" ��#��" $��" ��$"��%��&

• �� !��!��%��%�� &

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THE EARTH. MERIDIANS AND PARALLELS

� 1=Circle of latitude � 2=Meridian (geography)

A circle of latitude, on the Earth, is an imaginary east-west circle connecting all locations (not taking into account elevation) that share a given latitude. A location's position along a circle of latitude is given by its longitude.

Circles of latitude are often called parallels because they are parallel to each other. On some map projections, including the Equirectangular projection, they are drawn at equidistant intervals.

Circles of latitude become smaller the farther they are from the equator and the closer they are to the poles. A circle of latitude is perpendicular to all meridians at the points of intersection, and is hence a special case of a loxodrome.

Contrary to what might be assumed from their straight-line representation on some map projections, a circle of latitude is not, with the sole exception of the Equator, the shortest distance between two points lying on the Earth. In other words, circles of latitude (except for the Equator) are not great circles (see also great-circle distance). It is for this reason that an

airplane traveling between a European and North American city that share the same latitude will fly farther north, over Greenland for example.

Arcs of circles of latitude are sometimes used as boundaries between countries or regions where distinctive natural borders are lacking (such as in deserts), or when an artificial border is drawn as a "line on a map", as happened in Korea.

Longitude (λ)

Lines of longitude appear vertical with varying curvature in this projection; but are actually halves of great ellipses, with identical radii at a given latitude.

Latitude (φ)

Lines of latitude appear horizontal with varying curvature in this projection; but are actually circular with different radii. All locations with a given latitude are collectively referred to as a circle of latitude.

The equator divides the planet into aNorthern Hemisphere, a Southern Hemisphere and has a latitude of 0°.

Major circles of latitude

There are five major circles of latitude, listed below from north to south, with their values (Epoch 2010).[1]:

� Arctic Circle (66° 33 ′ 44″ N) � Tropic of Cancer (23° 26 ′ 16″ N) � Equator (0° latitude) � Tropic of Capricorn (23° 26 ′ 16″ S) � Antarctic Circle (66° 33 ′ 44″ S)

These circles of latitude (excluding the equator) mark the divisions between the five principal geographical zones.

A meridian (or line of longitude) is an imaginary arc on the Earth's surface from the North Pole to the South Pole that connects all locations running along it with a given longitude. The position of a point on the meridian is given by the latitude. Each meridian is perpendicular to all circles of latitudeat the intersection points. Each is also the same size, being half of a great circle on the Earth's surface and therefore measuring 20,003.93 km.

Since the meridian that passes through Greenwich, England, establishes the meaning of zero degrees of longitude, or the Prime Meridian, any other meridian is identified by the angle, referenced to the center of the earth as vertex, between where it and the prime meridian cross the equator. As there are 360 degrees in a circle, the meridian on the opposite side of the earth from Greenwich (which forms the other half of a circle with the two through Greenwich) is 180° longitude, and the others lie between 0° and 180° of West lo ngitude in the Western Hemisphere (west of Greenwich) and between 0° and 180° of Eas t longitude in the Eastern Hemisphere (east of Greenwich). Most maps show the lines of longitude.

The term "meridian" comes from the Latin meridies, meaning "midday"; the sun crosses a given meridian midway between the times of sunrise and sunset on that meridian. The same Latin stem gives rise to the terms a.m. (ante meridiem) and p.m. (post meridiem) used to disambiguate hours of the day when using the 12-hour clock.

The magnetic meridian is an equivalent imaginary line connecting the magnetic south and north poles and can be taken as the magnetic force lines along the surface of the earth[1]. That is, a compass needle will be parallel to the magnetic meridian. The angle between the magnetic and the true meridian is the Magnetic declination, which is relevant for navigating with a compass.[2]

INGLÉS ESPAÑOL A meridian (or line of longitude) Meridiano o línea de longitud A circle of latitude or paralel Círculo de latitud o paralelo zone huso Polar cap. Ice cap. Polar ice cap Casquete polar Zone (i.e. tropical zone) zona Time zones Husos horarios

EXERCISES (word problems)

1. Two cities are located on two meridians. The angle between these meridians is

225°. What will be the time difference?

2. Find the surface area of the Earth knowing that its radius is 6371 km.

3. Find the distance between two points on Earth, knowing that the geographic

coordinates are A(10°W, 25°S) and B(10°W,55°S). R=6371 Km.

4. The geographical coordinates of a town are (15°E, 45°N). Find the distance of this

town to the equator measured over the town’s meridian.

5. In geography, the antipodes of any place on Earth is the point on the Earth's surface

which is diametrically opposite to it. Two points that are antipodal to each other

are connected by a straight line running through the centre of the Earth. Find the

geographical coordinates of the antipodes of Rome (12° 40’ E, 41° 50’ N).

6. What are the geographical coordinates of the north and south poles?

7. Find the surface area of a time zone (un huso horario). R = 6371 km.

8. Calculate the distance traveled by an airplane flying along the common parallel from

a point in Europe with geographical coordinates of (8°E, 45°N) to another point in

America with coordinates (70°W, 45°N).

9. Two points A and B are located on the equator with longitudes of 20° E and 20° W.

What is the distance between them?. Length of equator=40030 km.

Sequences

Sequences

1. Number sequences

Consider the illustrated pattern of circles:

The first layer has just one blue ball.The second layer has three pink balls.The third layer has five black balls.The fourth layer has seven green balls.

If we let un represent the number of balls in the nth layer, then u1 = 1, u2 = 3, u3 = 5, and u4 = 7.

The pattern could be continued forever, generating the sequence of numbers:

1, 3, 5, 7, 9, 11, 13, 15, 17, ......

The string of dots indicates that the pattern continues forever.

A sequence is a set of terms, in a definite order, where the terms are obtained by some rule.

The sequence for the pattern of balls can be specified:

• using words: “The set of all odd numbers starting with 1”.

• using an explicit formula: un = 2n− 1 generates all terms. un is called the nth term or thegeneral term.

• using a recursive formula: un = 1 and un+1 = un + 2 for all n ≥ 1.

Check: u1 = 1

u2 = u1 + 2 = 1 + 2 = 3 X

u3 = u2 + 2 = 3 + 2 = 5 X

2. Arithmetic sequences

An arithmetic sequence is a sequence in which each term differs from the previous one by thesame fixed number. We call this number the common difference.

For example: 1, 5, 9, 13, 17, . . . is arithmetic as 5− 1 = 9− 5 = 13− 9, etc. The difference is 4.

Likewise, 42, 37, 32, 27, . . . is arithmetic as 37− 42 = 32− 37 = 27− 32, etc. The difference is −5.

In the sequence 1, 5, 9, 13, 17, . . . notice that

u1 = 1

u2 = 1 + 4 = 1 + 1 · 4u3 = 1 + 4 + 4+ = 1 + 2 · 4u4 = 1 + 4 + 4 + 4 = 1 + 3 · 4, etc.

This suggests that:

1

2 Sequences - 3 o ESO

If un is arithmetic then the nth term is un = u1 + (n− 1)d where u1 is the first term andd is the constant common difference.

Example 1

Consider the sequence 3, 9, 15, 21, 27, . . . :

a) Show that the sequence is arithmetic.

b) Find the formula for the general term un.

c) Find the 100th term of the sequence.

d) Is i) 489 ii) 1592 a member of the sequence?

a) 9− 3 = 6, 15− 9 = 6 21− 15 = 6, 27− 21 = 6

So, assuming that the pattern continues, consecutive terms differ by 6. Hence, the sequence isarithmetic with u1 = 3 and d = 6.

b) un = u1 + (n− 1)d ⇒ un = 3 + 6(n− 1) ⇒ un = 6n− 3

c) If n = 100, u100 = 6 · 100− 3 = 597.

d) i) Let un = 489 ⇒ 6n− 3 = 489 ⇒ n = 82

489 is a term of the sequence. In fact it is the 82nd term.

ii) Let un = 1592 ⇒ 6n− 3 = 1592 ⇒ n = 265 56

which is not possible as n is an integer. Hence 1592 cannot be a term.

Example 2

Find k given that k+5, −1 and 2k−1 are consecutive terms of an arithmetic sequence,and hence find the terms.

Since the terms are consecutive,

−1− (k + 5) = (2k − 1)− (−1) equating common differences

∴ −1− k − 5 = 2k − 1 + 1

∴ −k − 6 = 2k

∴ −6 = 3k

∴ k = −2

∴ the terms are 3, −1, −5.

Example 3

Find the general term un for an arithmetic sequence given that u3 = 4 and u7 = −24.

u7 − u3 = (u1 + 6d)− (u1 + 2d) = u1 + 6d− u1 − 2d = 4d

But u7 − u3 = −24− 4 = −28 ⇒ 4d = −28 ⇒ d = −7

Now u3 = u1 + 2 · (−7) ⇒ 4 = u1 − 14 ⇒ u1 = 18

Hence un = 18 + (n− 1)(−7) ⇒ un = 25− 7n


Sequences - 3 o ESO 3

Exercises - Set A

1. Consider the sequence 4, 11, 18, 25, 32, . . .

a) Show that the sequence is arithmetic. b) Find the formula for its general term.

c) Find its 30th term. d) Is 340 a member?

e) Is 738 a member?

2. Consider the sequence 67, 63, 59, 55, . . .

a) Show that the sequence is arithmetic. b) Find the formula for its general term.

c) Find its 60th term. d) Is −143 a member?

e) Is 85 a member?

3. An arithmetic sequence is defined by un = 11n− 7.

a) Find u1 and d. b) Find the 37th term.

c) What is the least term of the sequence which is greater than 250?

4. A sequence is defined by un =21− 4n

2.

a) Prove that the sequence is arithmetic. b) Find u1 and d. c) Find u55.

d) For what values of n are the terms of the sequence less than −300?

5. Find k given the consecutive arithmetic terms:

a) 31, k, 13 b) k, 8, k + 11 c) k + 2, 2k + 3, 17

6. Find the general term un for an arithmetic sequence given that:

a) u4 = 37 and u10 = 67.

b) u5 = −10 and u12 = −38.

c) the fourth term is −4 and the fifteenth term is 29.

d) the tenth and sixth terms are −16 and −13 respectively.

7. Consider the finite arithmetic sequence 3, 2 12 , 2 . . . ,−6.

a) Find u1 and d. b) How many terms does the sequence have?

8. An arithmetic sequence starts 17, 24, 31, 38, . . . What is the first term of the sequence toexceed 40 000?

3. Geometric sequences

A sequence is geometric if each term can be obtained from the previous one by multiplying bythe same non-zero constant.

For example: 2, 6, 18, 54, . . . is a geometric sequence as 2 · 3 = 6 and 6 · 3 = 18 and 18 · 3 = 54.

Notice thatu2 = u1 · 3u3 = u2 · 3 = u1 · 3 · 3 = u1 · 32

u4 = u3 · 3 = u1 · 32 · 3 = u1 · 33

u5 = u4 · 3 = u1 · 33 · 3 = u1 · 34



This suggests the following algebraic definition:

If un is geometric then un = u1 · rn−1 for all positive integers n.u1 is the first term and r is a constant called the common ratio.

Notice:

• r is called the common ratio becauseun+1

un

= r for all n.

• 2, 6, 18, 54, . . . is geometric with r = 3.

• 2,−6, 18,−54, . . . is geometric with r = −3.

Example 4

For the sequence 16, 8, 4, 2, 1 . . . :

a) Show that the sequence is geometric.

b) Find the general term un.

c) Hence, find the 10th term as a fraction.

a)8

16=

1

2,

4

8=

1

2,

2

4=

1

2,

1

2=

1

2

So, assuming the pattern continues, consecutive terms have a common ratio of1

2. Hence, the

sequence is geometric with u1 = 16 and r =1

2.

b) un = u1 · rn−1 ⇒ un = 16 ·(1

2

)n−1

= 24 · (2−1)n−1 = 24+(−n+1) = 25−n

c) u10 = 25−10 = 2−5 =1

25=

1

32

Example 5

k − 1, k + 2 and 3k are consecutive terms of a geometric sequence. Find k.

Equating common ratios gives3k

k + 2=

k + 2

k − 1⇒ 3k(k − 1) = (k + 2)2 ⇒

3k2 − 3k = k2 + 4k + 4 ⇒ 2k2 − 7k − 4 = 0 ⇒ k = 4 or k = −1

2

Check: If k = 4, the terms are: 3, 6, 12 (r = 2)

If k = −1

2, the terms are: −

3

2,3

2,−

3

2(r = −1)



Example 6

A geometric sequence has u2 = −5 and u5 = 40. Find its general term.

u2 = u1r = −5 and u5 = u1r4 = 40, so

u1r4

u1r=

40

−5⇒ r3 = −8 ⇒ r = −2

Then, u1(−2) = −5 ⇒ u1 =5

2, and un =

5

2· (−2)n−1

Exercises - Set B

1. For the geometric sequence with first two terms given, find b and c:

a) 3, 6, b, c, . . . b) 8, 2, b, c, . . . c) 15,−5, b, c, . . .

2. Consider the sequence 1, 3, 9, 27, . . .


b) Find un and hence find the 10th term.

3. Consider the sequence 40,−20, 10,−5 . . .


b) Find un and hence find the 12th term as a fraction.

4. Show that the sequence 16,−4, 1,−0, 25, . . . is geometric and hence find the 8th term as adecimal.

5. Find the general term of the geometric sequence: 3, 3√2, 6, 6

√2, . . .

6. Find k given that the following are consecutive terms of a geometric sequence:

a) k, 2, 6 b) 4, 6, k c) k, 2√2, k2

d) 3, k, 27 e) k, 3k, 10k + 7 f) k, k + 4, 8k + 2

7. Find the general term un of the geometric sequence which has:

a) u3 = 16 and u8 = 512 b) u3 = 32 and u6 = −4

c) u7 = 24 and u15 = 384 d) u3 = 3 and u9 =3

8

8. A geometric sequence has general term un with u3 = 12 and u7 =3

4. Find u12.

9. un is the general term of a geometric sequence.

a) If u2 = −2 12 and u5 =

5

16, find u10.

b) If u3 = 7 and u8 = −7, find u88.

c) If u3 = 18 and u5 = 162, find u11.



4. Sum of n terms of an arithmetic sequence

Recall that if the first term of an arithmetic sequence is u1 and the common difference is d, then theterms are: u1, u1 + d, u1 + 2d, u1 + 3d, etc.

Suppose that un is the last term of the sequence. Then, the sum of the n consecutive terms is:

Sn = u1 + (u1 + d) + (u1 + 2d) + · · ·+ (un − 2d) + (un − d) + un. But, reversing them:

Sn = un + (un − d) + (un − 2d) + · · ·+ (u1 + 2d) + (u1 + d) + u1

Adding these two expressions vertically, we get:

Sn = u1 + (u1 + d) + (u1 + 2d) + · · · + (un − 2d) + (un − d) + un

+ Sn = un + (un − d) + (un − 2d) + · · · + (u1 + 2d) + (u1 + d) + u1

2 · Sn = (u1 + un) + (u1 + un) + (u1 + un) + · · · + (u1 + un) + (u1 + un) + (u1 + un)︸︷︷︸

n of these

2 · Sn = n(u1 + un) ⇒ Sn =1

2n(u1 + un)

Example 7

Find the sum of the first 50 terms of the sequence 4, 7, 10, 13, . . .

The sequence is arithmetic with u1 = 4, d = 3 and n = 50. Then:

u50 = u1 + 49d = 4 + 49 · 3 = 151

S50 =1

2· 50(4 + 151) = 3875

Exercises - Set C

1. Find the sum of:

a) the first 20 terms of the sequence 3, 7, 11, 15, . . .

b) the first 22 terms of the sequence −6, 1, 8, 15, . . .

c) the first 40 terms of the sequence 100, 93, 86, 79, . . .

5. Sum of n terms of a geometric sequence

Let us consider the sum of the n first terms of a geometric sequence: Sn = u1+u2+ · · ·un−1+un

If we multiply both sides by the common ratio r:

r · Sn = u1 · r + u2 · r + · · ·+ un−1 · r + un · r , that is:

r · Sn = u2 + u3 + · · ·+ un + un · r

Subtracting: rSn − Sn = unr − u1 ⇒ Sn(r − 1) = unr − u1

Hence: Sn =un · r − u1

r − 1



Example 8

Find the sum of the first 12 terms of the sequence 2, 6, 18, 54, . . .

The sequence is geometric with u1 = 2, r = 3 and n = 12. Then:

u12 = 2 · 311 = 354 294

S50 =354 294 · 3− 2

3− 1=

1 062 880

2= 531 440

Exercises - Set D

1. Find the sum of:

a) the first 10 terms of the sequence 12, 6, 3, 1.5 . . .

b) the first 15 terms of the sequence 6,−3, 1.5,−0.75 . . .

Review Exercises

1. The first term of an arithmetic sequence is equal to 6 and the common difference is equal to3. Find a formula for the nth term and the value of the 50th term.

2. The first term of an arithmetic sequence is equal to 200 and the common difference is equalto −10. Find the value of the 20th term.

3. An arithmetic sequence has a common difference equal to 10 and its 6th term is equal to 52.Find its 15th term.

4. An arithmetic sequence has a its 5th term equal to 22 and its 15th term equal to 62. Findits 100th term.

5. Find the sum of all the integers from 1 to 1000.

6. Find the sum of the first 50 even positive integers.

7. Find the sum of all positive integers, from 5 to 1555 inclusive, that are divisible by 5.

8. Find the terms a2, a3, a4 and a5 of a geometric sequence if a1 = 10 and the common ratior = −1.

9. Find the 10th term of a geometric sequence if a1 = 45 and the common ration r = 0.2.

10. Find a20 of a geometric sequence if the first few terms of the sequence are given by −1/2,1/4 , −1/8 , 1/16 , . . .

11. Given the terms a10 = 3/512 and a15 = 3/16384 of a geometric sequence, find the exactvalue of the term a30 of the sequence.

12. Find the sum of the first 12 terms of the sequence: 1, 3, 9, 27 . . .

13. Find a20 given that a3 = 1/2 and a5 = 8.

14. Find a30 given that the first few terms of a geometric sequence are given by−2, 1,−1/2, 1/4 . . .

15. Find r given that a1 = 10 and a20 = 10−18.

✁ ✃ ✁ ✃ ✁ ✃ ✁ ✃ ✁ ✃ ✁ ✃ ✁ ✃


2nd term - ies jovellanos

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