INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 3 : Further Trig Algebraic Operations Quadratic...

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK

UNIT 3 :

FurtherTrig

AlgebraicOperations

Quadratic Functions

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UNIT 3 : AlgebraicOperations

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Please choose a question to attempt from the following:

1 2 3 4

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Unit 3 Menu

5 6 7 8

Express (m -7)

as a single fraction in its simplest form.

ALGEBRAIC OPERATIONS : Question 1

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4

2 ( 7)

m m

m

What would you like to do now?

Express (m -7)



EXIT

4

2 ( 7)

m m

m

Use the pattern

ab + c

dad + bc

bd=

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Express (m -7)



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4

2 ( 7)

m m

m

m2 - m

2m + 14=

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Begin Solution

Question 1

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Express

(m -7)

as a single fraction in its

simplest form.

4

2 ( 7)

m m

m

1. Use the pattern

ab + c

dad + bc

bd=

4

2 ( 7)

m m

m

= m (m + 7) - (2 x 4m)

2 (m + 7)

m2 + 7m - 8m2m + 14

=

m2 - m2m + 14

=Try another like this

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1. Use the pattern

ab + c

dad + bc

bd=

4

2 ( 7)

m m

m

m (m + 7) - (2 x 4m)

2 (m + 7)

m2 + 7m - 8m2m + 14

=

m2 - m2m + 14

=

To add or subtract fractions

use the results:

ab

+cd

ad + bcbd

=

ab

-cd

ad - bcbd

=

Try another

ALGEBRAIC OPERATIONS: Question 1B

Express (t 3)


7 4

10 ( 3)

t t

t

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Express (t 3)


7 4

10 ( 3)

t t

t

Use the pattern

ab + c

dad + bc

bd=

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Express (t 3)


7 4

10 ( 3)

t t

t

7t2 + 19t

10t – 30 =

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Question 1B

Express

(t 3)


simplest form.

1. Use the pattern

ab + c

dad + bc

bd=

7 4

10 ( 3)

t t

t

= 7t (t – 3) + (10 x 4t)

10 (t – 3)

7t2 – 21t +40t10t – 30

=

7t2 + 19t10t – 30

=

7 4

10 ( 3)

t t

t

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To add or subtract fractions

use the results:

ab

+cd

ad + bcbd

=

ab

-cd

ad - bcbd

=

1. Use the pattern

ab + c

dad + bc

bd=

7 4

10 ( 3)

t t

t

= 7t (t – 3) + (10 x 4t)

10 (t – 3)

7t2 – 21t +40t10t – 30

=

7t2 + 19t10t – 30

=Menu

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Comments

7t2 + 19t10t - 30

Note:

Always check that you have

cancelled as far as possible.

This is the final result,

it does not cancel further.

1. Use the pattern

ab + c

dad + bc

bd=

7 4

10 ( 3)

t t

t

= 7t (t – 3) + (10 x 4t)

10 (t – 3)

7t2 – 21t +40t10t – 30

=

7t2 + 19t10t – 30

=Menu

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ALGEBRAIC OPERATIONS: Question 2

2 2

3

2 15

5 4

a b

b a Express


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2 2

3

2 15

5 4

a b

b a Express


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Multiply top line then

bottom line.

Cancel numbers

then letters in alphabetical

order.


2 2

3

2 15

5 4

a b

b a Express


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3b2a=

Question 2

Express


simplest form.

2 2

3

2 15

5 4

a b

b a

2a2

5b x 15b2

4a3

30a2b2

20a3b=

1. Multiply top line then bottom line.

2. Cancel numbers then letters in alphabetical order.

3

2 a

b

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3b2a=


Comments

2a2

5b x 15b2

4a3

30a2b2

20a3b=

3b2a=



3

2 a

b

To multiply fractions use the result:

ab x c

dacbd=

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2a2

5b x 15b2

4a3

30a2b2

20a3b=

3b2a=



3

2 a

b

To simplify final answer write

out in full and cancel:

30a2b2

20a3b=

= 3b2a

30.a.a.b.b20.a.a.a.b

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2

3

2 6v v

w w Express


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Express


Multiply top line then

bottom line.

Cancel numbers

then letters in alphabetical

order.

2

3

2 6v v

w w

To divide by a fraction :

turn it upside down and multiply.

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Express


2

3

2 6v v

w w vw2

3=

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Question 2B

Express


simplest form.

1. To divide by a fraction : turn it upside down and multiply.


1

3

w2v

2

3

2 6v v

w w

2v2

w 6v w3

2v2

w x 6vw3

=


2v2w3

6vw=

vw2

3=Comments

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To divide fractions use the result:1. To divide by a fraction : turn it

upside down and multiply.


2v2

w 6v w3

2v2

w x 6vw3

=


2v2w3

6vw=

vw2

3=

1

3

w2v

ab ÷ c

dadbc= a

b x dc =

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1. To divide by a fraction : turn it upside down and multiply.


2v2

w 6v w3

2v2

w x 6vw3

=


2v2w3

6vw=

vw2

3=

1

3

w2v

To simplify final answer write

out in full and cancel:

2v2w3

6vw= 2.v.v.w.w.w

6.v.w

vw2

3=

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ALGEBRAIC OPERATIONS: Question 3 5 3

4 424 8d d

Simplify

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Reveal answer only

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Get hint


Deal with numbers and then apply laws

of indices: when dividing subtract

the powers.

Remember subtracting negative is like adding.


Simplify 5 3

4 424 8d d

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Simplify 5 3

4 424 8d d

= 3d2

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Simplify

.

Question 3 1. Deal with numbers and then apply

laws of indices: when dividing subtract the powers.

2. Remember subtracting negative is like adding.

5 3

4 424 8d d

24d5/4 8d–3/4

= 3d5/4-(–3/4)

= 3d8/4

= 3d2

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1. Deal with numbers and then apply laws of indices: when dividing subtract the powers.

2. Remember subtracting negative is like adding.

24d5/4 8d–3/4

= 3d5/4-(–3/4)

= 3d8/4

= 3d2

Learn Laws of Indices:

m n m na a a

5 23 3a a

e.g.

73a

5 2( )3 3a

5 23 3a

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ALGEBRAIC OPERATIONS: Question 3B 5 2

3 3

2

a a

a

Simplify

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Deal with top row first. Apply laws of indices: when dividing subtract the

powers & when multiplying add

powers.

Now divide remembering subtracting negative is like adding.


Simplify

5 2

3 3

2

a a

a

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Simplify

5 2

3 3

2

a a

a

= a3

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Simplify

.

Question 3B 1. Deal with top row first. Apply laws

of indices: when dividing subtract the powers & when multiplying add powers.

2. Now divide remembering subtracting negative is like adding.

5 2

3 3

2

a a

a

a-2

a5/3 x a–2/3

= a1-(-2)

= a5/3 –2/3 a-2

= a3/3

a-2

= a3 Comments

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m n m na a a 1. Deal with top row first. Apply laws



a-2

a5/3 x a–2/3

= a1-(-2)

= a5/3 –2/3 a-2

= a3/3

a-2

= a3

m n m na a a 5 2

3 3a a

e.g.

5 2( )3 3a

33a

1a

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m n m na a a 1. Deal with top row first. Apply laws



a-2

a5/3 x a–2/3

= a1-(-2)

= a5/3 –2/3 a-2

= a3/3

a-2

= a3

m n m na a a e.g.

2

a

a

1 ( 2)a 1 2a 3a

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1 5 4

3 3 3m m m

Simplify giving your

answer with positive indices.

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1 5 4

3 3 3m m m


answer with positive indices.

Deal with numbers and then apply laws

of indices: when dividing subtract

the powers.

Remember subtracting negative is like adding.

Go to full solution

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Reveal answer only

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1 5 4

3 3 3m m m


answer with positive indices. = m2 + 1

m

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Question 4 1. Multiply out brackets remembering

to apply laws of indices: when multiplying add the powers.

2. Negative powers become positive on bottom line.

1 5 4

3 3 3m m m

Simplify

giving your answer with

positive indices.

m1/3( m5/3 + m-4/3 )

= m6/3 + m-3/3

= m2 + m-1

= m2 + 1 m

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m n m na a a

e.g.

2

1

a

2a

1. Multiply out brackets remembering to apply laws of indices: when multiplying add the powers.

2. Negative powers become positive on bottom line.

m1/3( m5/3 + m-4/3 )

= m6/3 + m-3/3

= m2 + m-1

= m2 + 1 m

1mm

aa

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1 1 3

4 4 4w w w

Simplify

giving your answer without indices.

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Get hint


Simplify


Multiply out

brackets remembering to

apply laws of indices: when

multiplying add the powers.

Zero power is 1 and ½ power is

square root.

ALGEBRAIC OPERATIONS: Question 4B 1 1 3

4 4 4w w w

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Simplify


1 1 3

4 4 4w w w

= 1 - w

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Question 4B 1. Multiply out brackets remembering

to apply laws of indices: when multiplying add the powers.

2. Zero power is 1 and ½ power is square root.

Simplify


1 1 3

4 4 4w w w

w-1/4( w1/4 - w3/4 )

= w0 - w1/2

= 1 - w

= w-1/4+1/4 - w-1/4+3/4

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e.g.

1. Multiply out brackets remembering to apply laws of indices: when multiplying add the powers.

2. Zero power is 1 and ½ power is square root.

w-1/4( w1/4 - w3/4 )

= w0 - w1/2

= 1 - w

= w-1/4+1/4 - w-1/4+3/4

0

1

1

nn

a

a a

12

133

a a

a a

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Evaluate 7c3/4 when c = 16


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Deal with indices first. Power ¾ is 4th root cubed.

Evaluate root before

power


Evaluate 7c3/4 when c = 16


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Evaluate 7c3/4 when c = 16 = 56



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Question 5 1. Deal with indices first. Power ¾ is

4th root cubed.

= 56

Evaluate

7c3/4 when c = 16 c3/4 = (4c)3

= (416)3

= (2)3

= 8

So 7c3/4 = 7 x 8

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Comments


1. Deal with indices first. Power ¾ is 4th root cubed.

= 56

= (4c)3

= (416)3

= (2)3

= 8

So 7c3/4 = 7 x 8

m n mna a

Think “Flower Power”:

Power on top

Root at bottom

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1. Deal with indices first. Power ¾ is 4th root cubed.

= 56

= (4c)3

= (416)3

= (2)3

= 8

So 7c3/4 = 7 x 8

m n mna ae.g. 4 3 43

4 3 4 433

4

8 8 ( 8)

2 16

a a

Always find root before power

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Evaluate 10f -1/2 when f = 25


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Evaluate 10f -1/2 when f = 25

Deal with indices first. Power – ½ is square root on

bottom line.


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= 2 Evaluate 10f -1/2 when f = 25


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Question 5B

Evaluate

10f -1/2 when f = 25

1. Deal with indices first. Power – ½ is square root on bottom line.

10f -1/2

= 10 f

= 1025

= 10 5

= 2

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e.g.

1. Deal with indices first. Power – ½ is square root on bottom line.

10f -1/2

= 10 f

= 1025

= 10 5

= 2

1

1mm

nn

aa

a a

33

12

1 12

2 8

9 9 3

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f(x) = 7x1/2 . Find the value of x if f(x) = 63.


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Equate both things that are equal to

f(x).

To get rid of square roots square both

sides.




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X = 81




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Question 6 1. Equate both things that are equal to

f(x). f(x) = 7x1/2 .

Find the value of x

if f(x) = 63.

7x1/2 = 63

so x1/2 = 9

(7)

2. Remember power ½ is the square root.

so x = 9

3. Now square each side.

so x = 92

ie x = 81Comments

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Comments

Learn Laws of Indices:1. Equate both things that are equal

to f(x).

7x1/2 = 63

so x1/2 = 9

(7)

2. Remember power ½ is the square root.

so x = 9

3. Now square each side.

so x = 92

ie x = 81

12a a

Note:In solving the equation

Square both sides2

4

4

16

a

a

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Equate both things that are equal to

f(x).

To get rid of cube roots cube both

sides.




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X = 1000



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Question 6B 1. Equate both things that are equal to

f(x). f(x) = 2x1/3 .

Find the value of x

if f(x) = 20. 2. Remember power 1/3 is the cube

root.

3. Now cube each side.

ie x = 1000

2x1/3 = 20

so x1/3 = 10

(2)

so 3x = 10

so x = 103

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133a a

Note:In solving the equation

cube both sides

3 2a 32 8a

1. Equate both things that are equal to f(x).

2. Remember power 1/3 is the cube root.

3. Now cube each side.

ie x = 1000

2x1/3 = 20

so x1/3 = 10

(2)

so 3x = 10

so x = 103

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Simplify 75 - 27 + 48


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Get hint

Find the largest “perfect square” factor of each of

the numbers! Collect identical

surds in same way as you would

letters. .


Simplify 75 - 27 + 48


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Simplify 75 - 27 + 48 = 63



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Question 7 1. Find the largest “perfect square”

factor of each of the numbers!

2. Collect identical surds in same way as you would letters.

Simplify

75 - 27 + 48 = 25 x 3 - 9 x 3 + 16 x 3

= 53 - 33 + 43

= 63 5x – 3x + 4x = 6x

75 - 27 + 48

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1. Find the largest “perfect square” factor of each of the numbers!


= 25 x 3 - 9 x 3 + 16 x 3

= 53 - 33 + 43

= 63

5x – 3x + 4x = 6x

. .a b a b

e.g. 75 25.3

5 3

25. 3

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= 25 x 3 - 9 x 3 + 16 x 3

= 53 - 33 + 43

= 63

5x – 3x + 4x = 6x

The key to these simplificationquestions is that all of theindividual terms can be reduced to a multiple of the same basic surd.

So check that once you have taken “perfect squares” all terms have the same basic surd. If not you may not have used thehighest perfect square for a term.

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Simplify 80 + 45 - 180


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Simplify 80 + 45 - 180




letters. .


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Simplify 80 + 45 - 180 = 5



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Question 7B

Simplify

80 + 45 - 180



= 16 x 5 +

80 + 45 - 180

9 x 5 - 36 x 5

= 45 + 35 - 65

= 5 4x + 3x – 6x = x

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= 16 x 5 +

80 + 45 - 180

9 x 5 - 36 x 5

= 45 + 35 - 65

= 5

4x + 3x – 6x = x

. .a b a b

80 16.5

4 5

16. 5

e.g.

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Comments



= 16 x 5 +

80 + 45 - 180

9 x 5 - 36 x 5

= 45 + 35 - 65

= 5

4x + 3x – 6x = x

The key to these simplificationquestions is that all of theindividual terms can be reduced to a multiple of the same basic surd.

So check that once you have taken “perfect squares” all terms have the same basic surd. If not you may not have used thehighest perfect square for a term.

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ALGEBRAIC OPERATIONS: Question 7C

Simplify 75 - 27 + 48 300 - 12


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Get hint

Deal with top and bottom lines separately. For each, find the

largest “perfect square” factor of

each of the numbers!


Simplify 75 - 27 + 48 300 - 12

Now bring both parts together

again. Cancel out where the same basic surd is in

evidence on both top and bottom.


Go to full solution

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Simplify 75 - 27 + 48 300 - 12

3/4=


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Question 7C 1. Deal with top and bottom lines separately. For each, find the largest “perfect square” factor of each of the numbers!

5x - 3x + 4x = 6x

Simplify

75 - 27 + 48

300 - 12 = 25 x 3 -

75 - 27 + 48

= 53 - 33 + 43

= 63

Top Line:

9 x 3 + 16 x 3

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Question 7C 1. Deal with top and bottom lines seperately. For each, find the largest “perfect square” factor of each of the numbers!

10x - 2x = 8x

Simplify

75 - 27 + 48

300 - 12

Bottom Line:

300 - 12

= 100 x 3 -

= 103 - 23

= 83

4 x 3

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Question 7C 2. Now bring both parts together again. Cancel out where the same basic surd is in evidence on both top and bottom.

Simplify

75 - 27 + 48

300 - 12

75 - 27 + 48

300 - 12

6383

3/4

=

=

3

4

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. .a b a b

80 16.5

4 5

16. 5

e.g.

1. Deal with top and bottom lines seperately. For each, find the largest “perfect square” factor of each of the numbers!

5x - 3x + 4x = 6x

= 25 x 3 -

75 - 27 + 48

= 53 - 33 + 43

= 63

Top Line:

9 x 3 + 16 x 3

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Comments

In questions involving fractions always treat each line separately.Reduce each line to its simplest form before dividing.

2. Now bring both parts together again. Cancel out where the same basic surd is in evidence on both top and bottom.

75 - 27 + 48

300 - 12

6383

3/4

=

=

The key to these simplificationquestions is that all of theindividual terms can be reduced to a multiple of the same basic surd. In fractions these will thenoften cancel.

Remember to cancel numbers too!!

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12

6Express with a rational denominator.


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Get hint

To change the look of a fraction but

not the value multiply top & bottom by the

same amount ie 6.


12


a a a


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= 26

12




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Question 8 1. To change the look of a fraction but not the value multiply top & bottom by the same amount ie 6.12

6Express

with a rational

denominator.

12

6 =12

6x 6

6

126

6=

= 26

a a a

2

1

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Comments

Learn Laws of Surds:

Rationalising the denominator:

1

a Required to remove from the denominator.

Multiply top and bottom by a

1.a a

aa a

1. To change the look of a fraction but not the value multiply top & bottom by the same amount ie 6.

12

6 =12

6x 6

6

126

6=

= 26

a a a

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e.g.

Rationalising the denominator:1. To change the look of a fraction but not the value multiply top & bottom by the same amount ie 6.

12

6 =12

6x 6

6

126

6=

= 26

a a a

1 1 5 5.

55 5 5

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843

Simplify


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letters. .

843

Simplify


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= 27 843

Simplify



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Question 8B 1. Re-write expression using laws of surds

843

Simplify

= 27

843 =

84 3

= 28

= 4 x 7

a a

bb

. .a b a b

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Comments


e.g.

1. Re-write expression using laws of surds

= 27

843 =

84 3

= 28

= 4 x 7

. .a b a b

a a

bb

a a

bb

24 244 2

66

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Comments


e.g.

1. Re-write expression using laws of surds

= 27

843 =

84 3

= 28

= 4 x 7

. .a b a b

a a

bb

. .a b a b

75 25.3 25. 3 5 3

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Find the value of tanx° giving your answer as a fraction with a rational denominator.

x°

3m

62mWhat would you like to do now?

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Reveal answer only

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Get hint

To change the look of a fraction but not the value multiply top & bottom by the same

amount ie 2.


When dealing with fractions check that you have

simplified as far as possible.


x°

3m

62m

0tanopp

xadj


Go to full solution

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Reveal answer only

EXIT



x°

3m

62m

2

4=


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Question 8C

Find the value of tanx°

giving your answer as

a fraction with a

rational denominator.

x°

3m

62m


tanx° = 3

62

=3

62x 2

2

32

6x2=

32

12=

2

4=

0tanopp

xadj

Simplify!!!1

4

Comments

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Continue Solution

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Comments

Learn Laws of Surds:1. To change the look of a fraction

but not the value multiply top & bottom by the same amount ie 2.

tanx° = 3

62

=3

62x 2

2

32

6x2=

32

12=

2

4=

Rationalising the denominator:

1

a Required to remove from the denominator.

Multiply top and bottom by a

1.a a

aa a

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Comments


e.g.


tanx° = 3

62

=3

62x 2

2

32

6x2=

32

12=

2

4=

2 2 3 2 3.

33 3 3

In questions involving fractionsmake sure you have simplified as far as possible.

Simplify!!!

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UNIT 3 :Further Trig



1 2

EXITBack to

Unit 3 Menu

Further Trig : Question 1

The graph below shows a curve with equation in the form y = asinbx° .

Write down the values of a and b.

1200 2400 3600

y = asinbx°

EXIT

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Further Trig Menu

Reveal answer only

Get hint





1200 2400 3600

y = asinbx°

a = amplitude = ½ vertical extent. b = no. of waves in 3600

EXIT

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Further Trig Menu

Reveal answer only





1200 2400 3600

y = asinbx°

a = 2 b = 3

EXIT

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Further Trig Menu



Question 1

1200 2400 3600

y = asinbx°


1. a = amplitude = ½ vertical extent.

max /min = ±2 so a = 2

2. b = no. of waves in 3600

3 complete waves from 0 to 360

so b = 3

Comments

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Comments





so b = 3

Learn basic trig graphs:

i.e. y = sinx˚

Max. = 1Min. = -1One cycle in 360˚

360

-1

1

x

y

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Comments





so b = 3


i.e. y = sinx˚

360

-1

1

x

y

y = asinx˚

Stretch factor

y = sinbx˚

Number of cycles in 360˚

One cycle

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Further Trig : Question 1B

The graph below shows a curve with equation in the form y = acosbx° .


y = acosbx°

900 1800 36002700

EXIT

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Further Trig Menu

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y = acosbx°

900 1800 36002700

a = amplitude = ½ vertical extent. b = no. of waves in 3600

EXIT

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Further Trig Menu

Reveal answer only





y = acosbx°

900 1800 36002700

a = 5 b = 2

EXIT

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Further Trig Menu


y = acosbx°

900 1800 36002700

Question 1B






so b = 2

Comments

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so b = 2


i.e. y = cosx˚

Max. = 1Min. = -1One cycle in 360˚

360

-1

1

x

y

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360

-1

1

x

y

Comments





so b = 3


i.e. y = cosx˚

y = acosx˚

Stretch factor

y = cosbx˚

Number of cycles in 360˚

One cycle

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Solve 5sinx° - 1 = 1 where 0<x<360

EXIT

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Further Trig Menu

Reveal answer only

Get hint




Always re-arrange given equation to:

sin x =cosx =tan x=

Use CAST diagram to decide which quadrants

angles lie.

Remember that answers to trig equations come

in pairs.

EXIT

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Further Trig Menu

Reveal answer only




x = 23.6°

x = 156.4°

EXIT

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Further Trig Menu



Question 2 1. Always re-arrange to sinx0 = …...

2. Use the CAST diagram to decide which quadrant angles lie in.

Solve

5sinx° - 1 = 1

where 0<x<360

sinx = 0.4

5sinx° - 1 = 1

5sinx° = 2

180 -

180 + 360 -

sin all

tan cos

Q1 or Q2

Where is sin

positive?

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Question 2 1. Always re-arrange to sinx0 = …...

3. Calculate angles remembering that answers to trig equations come in pairs.

Solve

5sinx° - 1 = 1

where 0<x<360

sinx = 0.4

5sinx° - 1 = 1

5sinx° = 2

Q1 or Q2

sin-1 0.4 = 23.6

Q1: angle = 23.6°

Q2: angle = 180° - 23.6°

= 156.4°Comments

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Comments

We are finding the two angles at which the sine curve reaches a height of 0.4

90 180 270 360

-1

-0.5

0.5

1

x

y

0.4

23.6 156.4

1. Always re-arrange to sinx0 = …...


sinx = 0.4

5sinx° - 1 = 1

5sinx° = 2

Q1 or Q2

Q2: angle = 180° - 23.6°

sin-1 0.4 = 23.6

Q1: angle = 23.6°

= 156.4°Menu

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Solve 3tanx° + 7 = 2 where 0<x<360

EXIT

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Further Trig Menu

Reveal answer only

Get hint




Always re-arrange given equation to:

sin x =cosx =tan x=

Use CAST diagram to decide which quadrants

angles lie.

Remember that answers to trig equations come

in pairs.

EXIT

Go to full solution

Go to Comments

Further Trig Menu

Reveal answer only




x = 121.0°

x = 301.0°

EXIT

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Further Trig Menu


Question 2B

Solve

3tanx° + 7 = 2

where 0<x<360

1. Always re-arrange to tanx0 = …...

tanx = -5/3

3tanx° + 7 = 2

3tanx° = -5

2. Use the CAST diagram to decide which quadrant angles lie in.

180 -

180 + 360 -

sin all

tan cos

Q2 or Q4

Where is tan

negative?

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Question 2B

Solve

3tanx° + 7 = 2

where 0<x<360


tanx = -5/3

3tanx° + 7 = 2

3tanx° = -5

Q2 or Q4

tan-1 (53) = 59.0

Q2: angle = 180° - 59.0°

Q4: angle = 360° - 59.0°


= 301.0°

= 121.0°

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Comments


tanx = -5/3

3tanx° + 7 = 2

3tanx° = -5

tan-1 (53) = 59.0

Q2: angle = 180° - 59.0°

Q4: angle = 360° - 59.0°


= 301.0°

= 121.0°

We are finding the two angles at which the tan graph reaches a height of 5

3

90 180 270 360

-3

-2

-1

1

2

3

x

y

121.0 301.0

5

3

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Comments


tanx = -5/3

3tanx° + 7 = 2

3tanx° = -5

tan-1 (53) = 59.0

Q2: angle = 180° - 59.0°

Q4: angle = 360° - 59.0°


= 301.0°

= 121.0°

1 5tan

3x

Note:

Never put a negative value into the

inverse trig function

DO:

then CAST diagram

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UNIT 3 : Quadratic Functions



1 2

EXITBack to

Unit 3 Menu

3 4

Quadratic Functions: Question 1

Solve the equation x2 – 6x + 7 = 0 giving your answers to 2 decimal places.

EXIT

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Go to Quadratics Menu

Reveal answer only

Get hint




Write down values of a,

b & c. Evaluate b2 – 4ac .

Now use quadratic formula,

remembering that you should get

two answers.

Remember to round.

EXIT

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Reveal answer only




x = 4.41 or 1.59

EXIT

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Question 1

Solve the equation

x2 – 6x + 7 = 0

giving your answers

to 2 decimal places.

1. Write down values of a, b & c.

x2 – 6x + 7 = 0

a = 1 b = (-6) c = 7

2. Evaluate b2 – 4ac .

b2 – 4ac = (-6)2 – (4x1x7)

= 36 – 28

= 8

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Question 1

Solve the equation

x2 – 6x + 7 = 0

giving your answers



x2 – 6x + 7 = 0

a = 1 b = -6 c = 7

3. Now use quadratic formula.

x = -b ± (b2 – 4ac )

2a

x = 6 ± 8

24. Rewrite with brackets and now use

calculator.

= (6 + 8) 2 or (6 - 8) 2

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Question 1

Solve the equation

x2 – 6x + 7 = 0

giving your answers



x2 – 6x + 7 = 0

a = 1 b = -6 c = 7

4. Rewrite with brackets and now use calculator.

= (6 + 8) 2 or (6 - 8) 2 = 4.414… or 1.585..

5. Remember to round.

x = 4.41 or 1.59Comments

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Comments


x2 – 6x + 7 = 0

a = 1 b = -6 c = 7


x = -b ± (b2 – 4ac )

2a

x = 6 ± 8


calculator.

= (6 + 8) 2 or (6 - 8) 2

For any question involving a quadratic in which you are asked to give your answer to a given number of decimal places

OR

a given number of significant figures:

THINK QUADRATIC FORMULA!!

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Comments


x2 – 6x + 7 = 0

a = 1 b = -6 c = 7


x = -b ± (b2 – 4ac )

2a

x = 6 ± 8


calculator.

= (6 + 8) 2 or (6 - 8) 2

ax2 + bx + c = 0

Refer to the Formula Sheet:

x = -b ± (b2 – 4ac )

2a

Quadratic Formula:

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Comments


x2 – 6x + 7 = 0

a = 1 b = -6 c = 7


x = -b ± (b2 – 4ac )

2a

x = 6 ± 8


calculator.

= (6 + 8) 2 or (6 - 8) 2

Take care when allocating a

value to a, b and c.

ax2 + bx + c = 0

1x2 - 6x + 7 = 0

a =1b = (-6)c = 7

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Note:

Finding first eases working.

Bracket all negative numbers. Watch out for double negative!

Comments


x2 – 6x + 7 = 0

a = 1 b = -6 c = 7

b2 – 4ac


b2 – 4ac = (-6)2 – (4x1x7)

= 36 – 28

= 8

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Quadratic Functions: Question 1B

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Reveal answer only

EXIT

Get hint

Solve the equation 3x2 + 5x - 1 = 0 giving your answers to 2 significant figures.



Go to full solution

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Reveal answer only

EXIT





two answers.

Remember to round.




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EXIT

x = 0.18 or -1.8



Comments

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Continue Solution

Question 1B

Quadratics Menu

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3x2 + 5x – 1 = 0

a = 3 b = 5 c = (-1)


b2 – 4ac = (5)2 – (4 x 3 x -1)

= 25 – (- 12)

= 37

Solve the equation

3x2 + 5x - 1 = 0

giving your answers

to 2 significant figures.

Comments

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Continue Solution

Question 1B

Quadratics Menu

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x = -b ± (b2 – 4ac )

2a

x = -5 ± 37


calculator.

= (-5 + 37) 2 or (-5 - 37) 2

3x2 + 5x – 1 = 0

a = 3 b = 5 c = (-1)

Solve the equation

3x2 + 5x - 1 = 0

giving your answers


Comments

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Continue Solution

Question 1B

Quadratics Menu

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= (-5 + 37) 2 or (-5 - 37) 2


x = 0.18 or -1.8

3x2 + 5x – 1 = 0

a = 3 b = 5 c = (-1)

Solve the equation

3x2 + 5x - 1 = 0

giving your answers


= 0.180… or -1.847..


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OR





x = -b ± (b2 – 4ac )

2a

x = -5 ± 37

24. Rewrite with brackets and now

use calculator.

= (-5 + 37) 2 or (-5 - 37) 2

3x2 + 5x – 1 = 0

a = 3 b = 5 c = (-1)

Comments

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ax2 + bx + c = 0


x = -b ± (b2 – 4ac )

2a

Quadratic Formula:



x = -b ± (b2 – 4ac )

2a

x = -5 ± 37


calculator.

= (-5 + 37) 2 or (-5 - 37) 2

3x2 + 5x – 1 = 0

a = 3 b = 5 c = (-1)

Comments

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ax2 + bx + c = 0

3x2 + 5x - 1 = 0

a =3b = 5c = -1



x = -b ± (b2 – 4ac )

2a

x = -5 ± 37


calculator.

= (-5 + 37) 2 or (-5 - 37) 2

3x2 + 5x – 1 = 0

a = 3 b = 5 c = (-1)

Note:



Comments

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b2 – 4ac1. Write down values of a, b & c.

3x2 + 5x – 1 = 0

a = 3 b = 5 c = (-1)


b2 – 4ac = (5)2 – (4 x 3 x -1)

= 25 – (- 12)

= 37


Go to full solution

Go to Comments


Reveal answer only

EXIT

Get hint

Solve the equation 3x2 = x + 1 giving your answers to 2 decimal places.



Go to full solution

Go to Comments


Reveal answer only

EXIT





two answers.

Remember to round.


Re-write quadratic:

make it equal to

zero before solving.



Go to full solution

Go to Comments


EXIT


x = 0.77 or -0.43



Comments

Begin Solution

Question 2

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3x2 – x – 1 = 0

a = 3 b = (-1) c = (-1)


b2 – 4ac = (-1)2 – (4 x 3 x -1)

= 1 – (-12)

= 13Continue Solution

Solve the equation

3x2 = x + 1

giving your answers


1. Rearrange in quadratic form.

3x2 = x + 1

Comments

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Question 2

Quadratics Menu

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3x2 –x – 1 = 0

a = 3 b = (-1) c = (-1)


x = -b ± (b2 – 4ac )

2a

x = 1 ± 13


calculator.

= (1 + 13) 2 or (1 - 13) 2

Solve the equation

3x2 = x + 1

giving your answers


Comments

Begin Solution

Question 2

Quadratics Menu

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3x2 – x – 1 = 0

a = 3 b = (-1) c = (-1)


= (1 + 13) 2 or (1 - 13) 2


x = 0.77 or -0.43Try another like this

Solve the equation

3x2 = x + 1

giving your answers


= 0.767… or -0.434..


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OR




3x2 – x – 1 = 0

a = 3 b = (-1) c = (-1)


b2 – 4ac = (-1)2 – (4 x 3 x -1)

= 1 – (-12)

= 13


3x2 = x + 1

Comments

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You must put the quadratic into standard quadratic form (make equation equal zero) before attempting to solve.


3x2 – x – 1 = 0

a = 3 b = (-1) c = (-1)


b2 – 4ac = (-1)2 – (4 x 3 x -1)

= 1 – (-12)

= 13


3x2 = x + 1

Comments

Quadratics Menu

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ax2 + bx + c = 0


x = -b ± (b2 – 4ac )

2a

Quadratic Formula:


3x2 –x – 1 = 0

a = 3 b = (-1) c = (-1)


x = -b ± (b2 – 4ac )

2a

x = 1 ± 13


use calculator.

= (1 + 13) 2 or (1 - 13) 2

Comments

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ax2 + bx + c = 0

3x2 - 1x - 1 = 0

a =3b = (-1)c = (-1)


3x2 –x – 1 = 0

a = 3 b = (-1) c = (-1)


x = -b ± (b2 – 4ac )

2a

x = 1 ± 13


use calculator.

= (1 + 13) 2 or (1 - 13) 2

Note:



Comments

Quadratics Menu

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b2 – 4ac


3x2 – x – 1 = 0

a = 3 b = (-1) c = (-1)


b2 – 4ac = (-1)2 – (4 x 3 x -1)

= 1 – (-12)

= 13


3x2 = x + 1



Go to full solution

Go to Comments


Reveal answer only

EXIT

Get hint

Solve the equation 2x(x + 2) = 2 - x

giving your answers to 2 significant figures.



Go to full solution

Go to Comments


Reveal answer only

EXIT


b & c.

Evaluate b2 – 4ac .



two answers.

Remember to round.



Re-write quadratic: get rid of

brackets & make it equal to

zero before solving.



Go to full solution

Go to Comments


EXIT

x = 0.35 or -2.9




Comments

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Continue Solution

Question 2B

Quadratics Menu

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Solve the equation

2x(x + 2) = 2 - x

giving your answers



2x2 + 5x – 2 = 0

a = 2 b = 5 c = (-2)


b2 – 4ac = (5)2 – (4 x 2 x -2)

= 25 – (-16)

= 41


2x(x + 2) = 2 – x 2x2 + 4x – 2 + x= 0

Comments

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Continue Solution

Question 2B

Quadratics Menu

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x = -b ± (b2 – 4ac )

2a

x = -5 ± 41


calculator.

= (-5 + 41) 2 or (-5 - 41) 2

Solve the equation

2x(x + 2) = 2 - x

giving your answers


2x2 + 5x – 2 = 0

a = 2 b = 5 c = (-2)


Comments

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Continue Solution

Question 2B

Quadratics Menu

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= (-5 + 41) 2 or (-5 - 41) 2


x = 0.35 or -2.9

2x2 + 5x – 2 = 0

a = 2 b = 5 c = (-2)


Solve the equation

2x(x + 2) = 2 - x

giving your answers


= 0.350… or -2.850..


Comments

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OR




2x2 + 5x – 2 = 0

a = 2 b = 5 c = (-2)


b2 – 4ac = (5)2 – (4 x 2 x -2)

= 25 – (-16)

= 41


2x(x + 2) = 2 – x 2x2 + 4x – 2 + x= 0

Comments

Quadratics Menu

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Next Comment


2x2 + 5x – 2 = 0

a = 2 b = 5 c = (-2)


b2 – 4ac = (5)2 – (4 x 2 x -2)

= 25 – (-16)

= 41


2x(x + 2) = 2 – x 2x2 + 4x – 2 + x= 0

You must put the quadratic into standard quadratic form (make equation equal zero) before attempting to solve.

Comments

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ax2 + bx + c = 0


x = -b ± (b2 – 4ac )

2a

Quadratic Formula:


x = -b ± (b2 – 4ac )

2a

x = -5 ± 41


use calculator.

= (-5 + 41) 2 or (-5 - 41) 2

2x2 + 5x – 2 = 0

a = 2 b = 5 c = (-2)


Comments

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ax2 + bx + c = 0

2x2 + 5x - 2 = 0

a =2b = 5c = -2


x = -b ± (b2 – 4ac )

2a

x = -5 ± 41


use calculator.

= (-5 + 41) 2 or (-5 - 41) 2

2x2 + 5x – 2 = 0

a = 2 b = 5 c = (-2)


Note:



Comments

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b2 – 4ac


2x2 + 5x – 2 = 0

a = 2 b = 5 c = (-2)


b2 – 4ac = (5)2 – (4 x 2 x -2)

= 25 – (-16)

= 41


2x(x + 2) = 2 – x 2x2 + 4x – 2 + x= 0


Go to full solution

Go to Comments

Go to Quadratics MenuReveal answer

EXIT

Get hint

The diagram below shows an L-shaped plot of land with dimensions as given.

(x+4) m

x m

x m

(x+3) m

(a) Show that the total area is given by the expression x2 + 7x m2.

(b) Hence find the value of x when this area is 60 m2.



(x+4) m

x m

x m

(x+3) m




EXIT

Find the area of each rectangle..

Add these to find total area.

In (b) make expression from (a) =

60.

Make it equal to

zero, factorise &

solve.

Go to full solution

Go to Comments


Reveal answer



(x+4) m

x m

x m

(x+3) m




EXIT


x = 5m

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Question 3

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(x+4) m

x m

x m

(x+3) m

1. Find the area of each rectangle separately and add together to get total area.

Area rectangle = x(x+4)

= x2 + 4x m2

(a)1

2

1

Length rectangle = (x + 3)2

= 3m

– x

= 3x m2Area rectangle 2

Hence total area = (x2 + 4x) + 3x

= x2 + 7x m2


3m

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Question 3

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(x+4) m

x m

x m

(x+3) m

2. Use the expression for area given in part (a) and make it equal to 60.

If total area = 60 (b)

1

2

And total area = x2 + 7x

(b) Hence find the value of x

when this area is 60 m2.

then x2 + 7x = 60

x2 + 7x - 60 = 0

(x + 12)(x – 5) = 0

So (x + 12) = 0 or (x – 5) = 0

ie x = -12 or x = 5

Length must be 5m as negative

value not valid.

3. Make the equation equal to zero, factorise and solve.



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And total area = x2 + 7x then x2 + 7x = 60

x2 + 7x - 60 = 0

(x + 12)(x – 5) = 0

So (x + 12) = 0 or (x – 5) = 0

ie x = -12 or x = 5


value not valid.


The quadratic equation can also be solved by applyingthe quadratic formula:

ax2 + bx + c = 0


x = -b ± (b2 – 4ac )

2a

Quadratic Formula:

Finding first eases working.Bracket all negative numbers.

b2 – 4ac

Comments

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x2 + 7x - 60 = 0

(x + 12)(x – 5) = 0

So (x + 12) = 0 or (x – 5) = 0

ie x = -12 or x = 5


value not valid.




ax2 + bx + c = 0

1x2 + 7x - 60 = 0

a =1b = 7c = (-60)

Comments

Quadratics Menu

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x2 + 7x - 60 = 0

(x + 12)(x – 5) = 0

So (x + 12) = 0 or (x – 5) = 0

ie x = -12 or x = 5


value not valid.


Make sure that you understand when a

negative answer may not be

correct in context given,

e.g. negative lengths or areas.



Go to full solution

Go to Comments


Reveal answer

EXIT

Get hint

The diagram below shows plans of the foundations for a house and garage.

12m

7m house

x m

? mgarage

Planning regulations state that the length of the garage should be 4m longer than its width and its floor area should be 25% of the floor area of the house.


to meet this requirement.

(a)Find an expression for the area of the garage in terms of x.






EXIT

Area = length x

width. Use information

to find length of garage.

Find the numerical value of the

required area for the garage.

Use expression from (a) to

form a quadratic equation.

Make it equal to

zero, factorise &

solve.

Go to full solution

Go to Comments


Reveal answer


12m

7m house

x m

? mgarage




12m

7m house

x m

? mgarage






EXIT Go to full solution

Go to Comments Go to Quadratics Menu

= x2 + 4x m2

x = 3What would you like to do now?

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Question 3B

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(a)

12m

7m house

x m

? m

garage

the length of the garage should

be 4m longer than its width and

its floor area should be 25%

of the floor area of the house.

Length of garage = (x+4) m

So area = x(x+4)

= x2 + 4x m2

1. Area = length x width. Width is x but still need to find length.

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Question 3B

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2. Area = length x width. Find the required area and use your answer to (a) to establish quadratic equation.

(b)

12m

7m house

x m

? m

garage





Area house = 12 x 7 = 84m2

Area garage = 25% of 84m2

= 21m2

So x2 + 4x = 21

x2 + 4x - 21 = 0

(x + 7)(x – 3) = 0

(x + 7) = 0 or (x – 3) = 0


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Question 3B

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(b)

12m

7m house

x m

? m

garage





x2 + 4x - 21 = 0

(x + 7)(x – 3) = 0

(x + 7) = 0 or (x – 3) = 0


ie. x = -7 or x = 3

Length must be 3m as

negative value not valid.


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The quadratic equation can also be solved by applyingthe quadratic formula:

ax2 + bx + c = 0


x = -b ± (b2 – 4ac )

2a

Quadratic Formula:

Finding first eases working.Bracket all negative numbers.

b2 – 4ac




= 21m2

So x2 + 4x = 21

x2 + 4x - 21 = 0

(x + 7)(x – 3) = 0

(x + 7) = 0 or (x – 3) = 0


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ax2 + bx + c = 0

1x2 + 4x - 21 = 0

a =1b = 4c = (-21)




= 21m2

So x2 + 4x = 21

x2 + 4x - 21 = 0

(x + 7)(x – 3) = 0

(x + 7) = 0 or (x – 3) = 0


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Make sure that you understand when a

negative answer may not be

correct in context given,

e.g. negative lengths or areas.

(b)

x2 + 4x - 21 = 0

(x + 7)(x – 3) = 0

(x + 7) = 0 or (x – 3) = 0


ie. x = -7 or x = 3

Length must be 3m as

negative value not valid.

(a) State the coordinates of E.

(b) The second parabola cuts the X-axis at C & D. D is the point (8,0), find the coordinates of C.

(c) Find the equation of the parabola with minimum turning point G.


Go to full solution

Go to CommentsReveal answer

EXIT Get hint

The blades in electric hair-clippers have grooves in the form of identical adjacent parabolas.

The second parabola has turning point E and equation y = (x – 5)2 – 9 .

X

Y

E F G

C D





In ‘completed square’, a

minimum value occurs when

bracket is zero.


EXIT

Use symmetry of a parabola to find other root if you know one.

Go to full solution

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Reveal answer


X

Y

E F G

C D

roots are equidistant from axis of symmetry.

use ‘completed square’ form to write equation of parabola.







X

Y

E F G

C D


EXIT


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E is the point (5,-9).

C is (2,0).

y = (x – 17)2 - 9

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Question 4

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1. In ‘completed square’, a minimum value occurs when bracket is zero.

X

Y

E F G

C D

The second parabola has

turning point E and equation

y = (x – 5)2 – 9 .

(a)

y will have a minimum value of

-9 when (x – 5)2 = 0.

ie when x = 5

so E is the point (5,-9).

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Question 4

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2. Use symmetry of a parabola to find other root if you know one.

X

Y

E F G

C D


(b) D is the point (8,0),

find the coordinates of C.

(b) Horizontal dist from E to D

= horizontal dist from E to C = 3 units

D has x-coordinate 8

so x-coordinate of C is 6 units less ie C is (2,0).

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Question 4

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3. Use symmetry of a parabola to find other turning points if you know one.X

Y

E F G

C D


(c)Distance from E to G

is 12 units so G is (17,-9)

4. If you know turning point, use ‘completed square’ form to write equation of parabola.

Equation of this parabola

in same form as the second

ie y = (x – 17)2 - 9Try another like this

(c) Find the equation of the

parabola with minimum G.


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Learn the following Results:

When a quadratic equation is in

the form y = (x – a)2 + b

the parabola has a

minimum T.P. at (a,b).

When a quadratic equation is in the form y = b - (x – a)2 the parabola has a maximum T.P. at (a,b).


(a)


-9 when (x – 5)2 = 0.

ie when x = 5


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y = (x – 5)2 - 9

Minimum T.P. at P(5,-9)

e.g.

P


(a)


-9 when (x – 5)2 = 0.

ie when x = 5


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P

The roots are equidistant from theturning point (or axis of symmetry)in the horizontal direction.

2. Use symmetry of a parabola to find other root if you know one.

(b) Horizontal dist from E to D

= horizontal dist from E to C = 3 units

D has x-coordinate 8

so y-coordinate of C is 6 units less ie C is (2,0).


Go to full solution


EXIT Get hint

The top of an ornate fence consists of a series of congruent adjacent parabolas.

The first parabola has turning point U and equation y = 4 – (x – 3)2 .

(a) State the coordinates of U.

(b) If T is the point (5,0) then find the coordinates of V, the maximum turning point on the second parabola.

(c) Find the equation of the parabola with maximum turning point W.

U V W

X

YT

The top of an ornate fence consists of a series of adjacent parabolas.





U V W

X

YT



Go to Comments


Reveal answer

In ‘completed square’, a

minimum value occurs when

bracket is zero.

Use symmetry of a parabola to find another tp if you know one.







EXIT


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U is the point (3,4).

V is (7,4).

y = 4 – (x – 11)2

The top of an ornate fence consists of a series of adjacent parabolas.


U V W

X

YT

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Question 4B

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1. In ‘completed square’, a maximum value occurs when bracket is zero.

(a)

y will have a maximum value of

4 when (x – 3)2 = 0.

ie when x = 3

so U is the point (3,4).

The first parabola has turning

point U and equation

y = 4 – (x – 3)2 .

U V W

X

YT

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Question 4B

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2. Use symmetry of a parabola to find another turning point if you know one.

(b) T is the point (5,0),

find the coordinates of V.

(b) Horizontal dist from U to T

= horizontal dist from T to V = 2 units

T has x-coordinate 5

so x-coordinate of V is 2 units more

ie V is (7,4).

U V W

X

YT


For identical parabolae,maximum values same so y-coordinate = 4

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Question 4B

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3. Use symmetry of a parabola to find other turning points if you know one.

(c)Distance from U to W

is 8 units so W is (11,4)



in same form as the first

ie y = 4 - (x – 11)2Try another like this


parabola with maximum W.

U V W

X

YT



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the parabola has a




(a)


4 when (x – 3)2 = 0.

ie when x = 3


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y = 4 – (x – 3)2

Maximum T.P. at P(3,4)

e.g.

P


(a)


4 when (x – 3)2 = 0.

ie when x = 3



(a) State the coordinates of G.

(b) Find the coordinates of H and J.

(c) Find the equation of the parabola with minimum turning point at F.

Quadratic Functions: Question 4C

Go to full solution


EXIT Get hint

The diagram below shows some identical adjacent parabolas.

The last parabola has a minimum turning point at G and cuts the x-axis at H and J.

Its equation is y = (x – 12)2 – 4 .

X

Y

F G

H J





Its equation is y = (x – 12)2 – 4 . In ‘completed

square’, a minimum value

occurs when bracket is zero.


EXIT

If not given a root you must solve the quadratic.

Go to full solution

Go to Comments


Reveal answer



X

Y

F G

H J







Go to Comments


G is the point (12,-4).

y = (x – 8)2 - 4




X

Y

F G

H J

H is (10,0) and J is (14,0)


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Question 4C

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(a)


-4 when (x – 12)2 = 0.

ie when x = 12

so G is the point (12,-4).

X

Y

F G

H J


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Question 4C

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2. If you are not given one of the roots you must solve quadratic to find roots.


X

Y

F G

H J

(b) Find the coordinates of

H and J.

(b) At H & J

(x – 12)2 – 4 = 0

so (x – 12)2 = 4

ie x – 12 = -2 or 2

ie x = 10 or 14

H is (10,0) and J is (14,0)

From diagram:

y = (x – 12)2 – 4 .

X

Y

F G

H J

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Question 4C

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3. Use symmetry of a parabola to find other turning points if you know one.

(c)Distance from G to H

is 2 units so F is 4 units horizontally from G.



in same form as the last


parabola with minimum F.


10 14

So F is the point (8,-4).

ie y = (x – 8)2 - 4

Continue Solution


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the parabola has a




(a)


-4 when (x – 12)2 = 0.

ie when x = 12


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y = (x – 12)2 – 4

Minimum T.P. at P(12,-4)

e.g.

P


(a)


-4 when (x – 12)2 = 0.

ie when x = 12


End of Unit 3

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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 3 : Further Trig Algebraic Operations Quadratic...

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