INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 3 :
Algebraic
Operations
Further
Trig
EXIT
Quadratic
Functions
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
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UNIT 3 :
Algebraic
Operations
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ALGEBRAIC OPERATIONS : Question 1
Express
m
4m

2 (m  7)
(m  -7)
as a single fraction in its simplest form.
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ALGEBRAIC OPERATIONS : Question 1
Express
m
4m

2 (m  7)
(m  -7)
Use the pattern
a + c = ad + bc
b d
bd
as a single fraction in its simplest form.
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ALGEBRAIC OPERATIONS : Question 1
Express
m
4m

2 (m  7)
(m  -7)
m2 - m
=
2m + 14
as a single fraction in its simplest form.
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Question 1
Express
m
4m

2 (m  7)
1. Use the pattern
a + c = ad + bc
b d
bd
m
4m

2 (m  7)
(m  -7)
as a single fraction in its
simplest form.
=
m (m + 7) - (2 x 4m)
2 (m + 7)
m2 + 7m - 8m
=
2m + 14
Begin Solution
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m2 - m
=
2m + 14
Comments
1. Use the pattern
To add or subtract fractions
use the results:
a + c = ad + bc
b d
bd
m
4m

2 (m  7)
a c
ad + bc
+
=
b d
bd
a c
ad - bc
=
b d
bd
m (m + 7) - (2 x 4m)
2 (m + 7)
m2 + 7m - 8m
=
2m + 14
m2 - m
=
2m + 14
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ALGEBRAIC OPERATIONS: Question 1B
Express
7t
4t

10 (t  3)
(t  3)
as a single fraction in its simplest form.
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ALGEBRAIC OPERATIONS: Question 1B
Express
7t
4t

10 (t  3)
(t  3)
Use the pattern
a + c = ad + bc
b d
bd
as a single fraction in its simplest form.
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ALGEBRAIC OPERATIONS: Question 1B
Express
7t
4t

10 (t  3)
(t  3)
7t2 + 19t
=
10t – 30
as a single fraction in its simplest form.
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Question 1B
Express
1. Use the pattern
7t
4t

10 (t  3)
a + c = ad + bc
b d
bd
7t
4t

10 (t  3)
(t  3)
as a single fraction in its
simplest form.
Begin Solution
Continue Solution
=
7t (t – 3)
+ (10 x 4t)
10 (t – 3)
7t2 – 21t +40t
=
10t – 30
7t2 + 19t
=
10t – 30
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Comments
To add or subtract fractions
1. Use the pattern
use the results:
a + c = ad + bc
b d
bd
a c
ad + bc
+
=
b d
bd
7t
4t

10 (t  3)
=
7t (t – 3)
a c
ad - bc
=
b d
bd
+ (10 x 4t)
10 (t – 3)
7t2 – 21t +40t
=
10t – 30
7t2 + 19t
=
10t – 30
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Comments
Note:
1. Use the pattern
Always check that you have
a + c = ad + bc
b d
bd
cancelled as far as possible.
This is the final result,
7t
4t

10 (t  3)
=
7t (t – 3)
+ (10 x 4t)
10 (t – 3)
it does not cancel further.
7t2 + 19t
10t - 30
7t2 – 21t +40t
=
10t – 30
7t2 + 19t
=
10t – 30
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ALGEBRAIC OPERATIONS: Question 2
Express
2a 2 15b2
 3
5b 4a
as a single fraction in its simplest form.
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ALGEBRAIC OPERATIONS: Question 2
Express
2a 2 15b2
 3
5b 4a
as a single fraction in its simplest form.
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Multiply top
line then
bottom line.
Cancel
numbers
then letters in
alphabetical
order.
ALGEBRAIC OPERATIONS: Question 2
Express
2a 2 15b2
 3
5b 4a
=
3b
2a
as a single fraction in its simplest form.
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Question 2
Express
1. Multiply top line then bottom line.
2a 2 15b2
 3
5b 4a
as a single fraction in its
simplest form.
2a2 x 15b2
5b
4a3
3
2b2
30a
= 20a3b
2
b
a
2. Cancel numbers then letters in
alphabetical order.
=
3b
2a
Begin Solution
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Comments
1. Multiply top line then bottom line.
2a2 x 15b2
5b
4a3
3
2b2
30a
= 20a3b
2
To multiply fractions use the
result:
axc
b d
=
ac
bd
b
a
2. Cancel numbers then letters in
alphabetical order.
=
3b
2a
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Comments
To simplify final answer write
1. Multiply top line then bottom line.
2a2 x 15b2
5b
4a3
3
2b2
30a
= 20a3b
2
out in full and cancel:
30a2b2
20a3b
=
30.a.a.b.b
20.a.a.a.b
b
a
=
3b
2a
2. Cancel numbers then letters in
alphabetical order.
=
3b
2a
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ALGEBRAIC OPERATIONS: Question 2B
Express
2v 2 6v
 3
w w
as a single fraction in its simplest form.
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ALGEBRAIC OPERATIONS: Question 2B
Express
2v 2 6v
 3
w w
as a single fraction in its simplest form.
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To divide by a
fraction :
turntop
it
Multiply
upside
down
line
then
and multiply.
bottom
line.
Cancel
numbers
then letters in
alphabetical
order.
ALGEBRAIC OPERATIONS: Question 2B
Express
2v 2 6v
 3
w w
=
vw2
3
as a single fraction in its simplest form.
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Question 2B
Express
2
2v
6v
 3
w w
as a single fraction in its
1. To divide by a fraction : turn it
upside down and multiply.
2v2
6v
w  w3
2
3
2v
w
= w x 6v
simplest form.
2. Multiply top line then bottom line.
1
=
v
w2
2
3
2v w
6vw
3
Begin Solution
Continue Solution
Comments
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3. Cancel numbers then letters in
alphabetical order.
=
vw2
3
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Comments
1. To divide by a fraction : turn it
upside down and multiply.
2v2
6v

w
w3
To divide fractions use the
result:
a÷c
b d
=
axd
b c
=
ad
bc
2
3
2v
w
=
x
w
6v
2. Multiply top line then bottom line.
1
=
v
w2
2
3
2v w
6vw
3
3. Cancel numbers then letters in
alphabetical order.
=
vw2
3
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Comments
To simplify final answer write
1. To divide by a fraction : turn it
upside down and multiply.
out in full and cancel:
2v2
6v

w
w3
=
2v2
w
x
w3
=
=
6v
2. Multiply top line then bottom line.
1
2v2w3
6vw
v
=
2.v.v.w.w.w
6.v.w
vw2
3
w2
2
3
2v w
6vw
3
3. Cancel numbers then letters in
alphabetical order.
=
vw2
3
Next Comment
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ALGEBRAIC OPERATIONS: Question 3
Simplify
5
4
24d  8d

3
4
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ALGEBRAIC OPERATIONS: Question 3
Simplify
5
4
24d  8d

3
4
Deal with numbers
and then apply laws
of indices: when
dividing subtract
the powers.
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EXIT
Remember
subtracting
negative is
like adding.
ALGEBRAIC OPERATIONS: Question 3
Simplify
5
4
24d  8d

3
4
= 3d2
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Question 3
Simplify
5
4
24d  8d
3

4
1. Deal with numbers and then apply
laws of indices: when dividing
subtract the powers.
24d5/4  8d–3/4
.
= 3d5/4-(–3/4)
2. Remember subtracting negative is
like adding.
= 3d8/4
= 3d2
Begin Solution
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Comments
Learn Laws of Indices:
1. Deal with numbers and then apply
laws of indices: when dividing
subtract the powers.
24d5/4  8d–3/4
= 3d5/4-(–3/4)
2. Remember subtracting negative is
like adding.
= 3d8/4
=
3d2
n
mn
5
2
a a  a
m
e.g.
a 3 a
a
3
5 (  2 )
3
3
a
5 2
3 3
a
7
3
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ALGEBRAIC OPERATIONS: Question 3B
Simplify
5
3
a a
2
a

2
3
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ALGEBRAIC OPERATIONS: Question 3B
Simplify
5
3
a a
2
a

2
3
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Reveal answer only
Go to full solution
Deal with top
row first. Apply
laws of indices:
when dividing
subtract the
powers & when
Now divide
multiplying add
remembering
powers.
subtracting
negative is
like adding.
Go to Comments
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ALGEBRAIC OPERATIONS: Question 3B
Simplify
5
3
a a
2
a

2
3
= a3
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Question 3B
5
3
Simplify
a a
a 2

2
3
1. Deal with top row first. Apply laws of
indices: when dividing subtract the
powers & when multiplying add
powers.
a5/3 x a–2/3
a-2
.
=
=
Begin Solution
Continue Solution
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a5/3 –2/3
a-2
a3/3
a-2
2. Now divide remembering
subtracting negative is like adding.
=
a1-(-2)
=
a3
Comments
1. Deal with top row first. Apply laws
of indices: when dividing subtract
the powers & when multiplying
add powers.
a5/3 x a–2/3
a-2
=
=
a5/3 –2/3
a-2
a3/3
a-2
2. Now divide remembering
subtracting negative is like
adding.
=
a1-(-2)
Learn Laws of Indices:
a a  a
m
n
mn
a a  a
m
e.g.
n
5
a a
a
3
m n
2
3
5 (  2 )
3
3
a
3
3
a
1
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=
a3
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Comments
1. Deal with top row first. Apply laws
of indices: when dividing subtract
the powers & when multiplying
add powers.
a5/3 x a–2/3
a-2
=
=
a5/3 –2/3
a-2
a3/3
a-2
2. Now divide remembering
subtracting negative is like
adding.
=
a1-(-2)
Learn Laws of Indices:
a a  a
m
n
mn
a a  a
a
e.g.
2
a
m
n
m n
1( 2)
a
1 2
a
a
3
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=
a3
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ALGEBRAIC OPERATIONS: Question 4
Simplify
4
 
 53
m m  m 3 


1
3
giving your
answer with positive indices.
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ALGEBRAIC OPERATIONS: Question 4
Simplify
4
 
 53
m m  m 3 


1
3
answer with positive indices.
giving your
Deal with numbers
and then apply laws
of indices: when
dividing subtract
the powers.
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EXIT
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Remember
subtracting
negative is
like adding.
ALGEBRAIC OPERATIONS: Question 4
Simplify
4
 
 53
m m  m 3 


1
3
answer with positive indices.
giving your
= m2 + 1
m
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Question 4
Simplify


m m  m 


1
3
5
3
giving your answer with
positive indices.
4

3
1. Multiply out brackets remembering
to apply laws of indices: when
multiplying add the powers.
m1/3( m5/3 + m-4/3 )
= m6/3 + m-3/3
= m2 + m-1
2. Negative powers become positive
on bottom line.
Begin Solution
= m2 + 1
m
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Comments
Learn Laws of Indices:
1. Multiply out brackets
remembering to apply laws of
indices: when multiplying add the
powers.
m1/3( m5/3 + m-4/3 )
= m6/3 + m-3/3
= m2 + m-1
2. Negative powers become positive
on bottom line.
= m2 + 1
m
mn
a a  a
1
m
a  m
a
m
e.g.
n
a 2
1
 2
a
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ALGEBRAIC OPERATIONS: Question 4B
Simplify
3
 14

4
w w w 



1
4
giving your answer without indices.
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ALGEBRAIC OPERATIONS: Question 4B
Simplify
3
 14

4
w w w 



1
4
Multiply out
brackets
remembering to
apply laws of
indices: when
giving your answer without indices.
multiplying
add
the
Zero
power
powers.
is 1 and ½
power is
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square root.
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ALGEBRAIC OPERATIONS: Question 4B
Simplify
3
 14

4
w w w 



1
4
= 1 - w
giving your answer without indices.
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Question 4B
Simplify


w w w 


1

4
1
4
3
4
giving your answer without
indices.
1. Multiply out brackets remembering
to apply laws of indices: when
multiplying add the powers.
w-1/4( w1/4 - w3/4 )
= w-1/4+1/4 - w-1/4+3/4
= w0 - w1/2
2. Zero power is 1 and ½ power is
square root.
Begin Solution
Continue Solution
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= 1 - w
Comments
Learn Laws of Indices:
1. Multiply out brackets
remembering to apply laws of
indices: when multiplying add the
powers.
-1/4
w
1 /4
(w
3 /4
-w
)
a0  1
a
1
n
na
= w-1/4+1/4 - w-1/4+3/4
= w0 - w1/2
2. Zero power is 1 and ½ power is
square root.
e.g.
a
a
1
1
2
 a
3
3a
= 1 - w
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ALGEBRAIC OPERATIONS: Question 5
Evaluate
7c3/4 when c = 16
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ALGEBRAIC OPERATIONS: Question 5
Evaluate
7c3/4 when c = 16
Deal with indices
first. Power ¾ is
4th root cubed.
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Evaluate
root before
power
ALGEBRAIC OPERATIONS: Question 5
Evaluate
7c3/4 when c = 16
= 56
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Question 5
Evaluate
7c3/4 when c = 16
1. Deal with indices first. Power ¾ is
4th root cubed.
c3/4 = (4c)3
= (416)3
= (2)3
= 8
So 7c3/4 = 7 x 8
Begin Solution
= 56
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Comments
Learn Laws of Indices:
1. Deal with indices first. Power ¾ is
4th root cubed.
a
m
n
 a
n
m
= (4c)3
= (416)3
= (2)3
Think “Flower Power”:
Power on top
= 8
3 /4
So 7c
= 7x8
= 56
Root at bottom
Next Comment
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Comments
Learn Laws of Indices:
1. Deal with indices first. Power ¾ is
4th root cubed.
= (4c)3
= (416)3
= (2)3
= 8
So 7c3/4 = 7 x 8
= 56
a
e.g.
m
a
8
4
4
n
 a
n
m
3
 3 a4
3
 3 84  ( 3 8) 4
 24  16
Always find
root before power
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ALGEBRAIC OPERATIONS: Question 5B
Evaluate
10f -1/2 when f = 25
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ALGEBRAIC OPERATIONS: Question 5B
Evaluate
10f -1/2 when f = 25
Deal with indices
first. Power – ½
is square root on
bottom line.
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ALGEBRAIC OPERATIONS: Question 5B
Evaluate
10f -1/2 when f = 25
=
2
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Question 5B
1. Deal with indices first. Power – ½ is
square root on bottom line.
Evaluate
10f -1/2 when f = 25
Begin Solution
10f -1/2
= 10
f
= 10
25
= 10
5
= 2
Continue Solution
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Comments
Learn Laws of Indices:
1. Deal with indices first. Power – ½
is square root on bottom line.
10f -1/2
= 10
f
= 10
25
= 10
5
= 2
a
m
1
 m
a
1
na
a
e.g.
n
1 1
2  3
2
8
3
9
1
2
 9 3
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ALGEBRAIC OPERATIONS: Question 6
f(x) = 7x1/2 .
Find the value of x if f(x) = 63.
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ALGEBRAIC OPERATIONS: Question 6
f(x) = 7x1/2 .
Find the value of x if f(x) = 63.
Equate both things
that are equal to
f(x).
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To get rid of
square roots
square both
sides.
ALGEBRAIC OPERATIONS: Question 6
f(x) = 7x1/2 .
Find the value of x if f(x) = 63.
X = 81
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Question 6
f(x) = 7x1/2 .
1. Equate both things that are equal to
f(x).
7x1/2 = 63
Find the value of x
(7)
x 1 /2 = 9
so
if f(x) = 63.
2. Remember power ½ is the square
root.
so
x = 9
3. Now square each side.
Begin Solution
so
x = 92
ie
x = 81
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Comments
Learn Laws of Indices:
1. Equate both things that are equal
to f(x).
a
7x1/2 = 63
so
(7)
x1/2 = 9
2. Remember power ½ is the square
root.
so
1
2
 a
Note:
In solving the equation
Square both sides
a 4
a  42
 16
x = 9
3. Now square each side.
so
x = 92
ie
x = 81
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ALGEBRAIC OPERATIONS: Question 6B
f(x) = 2x1/3 .
Find the value of x if f(x) = 20.
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ALGEBRAIC OPERATIONS: Question 6B
f(x) = 2x1/3 .
Find the value of x if f(x) = 20.
Equate both things
that are equal to
f(x).
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Reveal answer only
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To get rid of
cube roots
cube both
sides.
ALGEBRAIC OPERATIONS: Question 6B
f(x) = 2x1/3 .
Find the value of x if f(x) = 20.
X = 1000
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Question 6B
f(x) = 2x1/3 .
1. Equate both things that are equal to
f(x).
2x1/3 = 20 (2)
Find the value of x
x1/3 = 10
so
if f(x) = 20.
2. Remember power 1/3 is the cube
root.
so
3x =
10
3. Now cube each side.
Begin Solution
so
x = 103
Comments
ie
x = 1000
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Comments
1. Equate both things that are equal
to f(x).
2x1/3 = 20 (2)
x1/3 = 10
so
2. Remember power 1/3 is the cube
root.
so
3x =
Learn Laws of Indices:
a
1
3
3a
Note:
In solving the equation
cube both sides
3
a 2
a  23  8
10
3. Now cube each side.
so
x = 103
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ie
x = 1000
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ALGEBRAIC OPERATIONS: Question 7
Simplify
75 - 27 + 48
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ALGEBRAIC OPERATIONS: Question 7
Simplify
75 - 27 + 48
What would you like to do now?
Find the largest
“perfect square”
factor of each of
the numbers!
Collect identical
surds in same
way as you would
letters.
.
Reveal answer only
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ALGEBRAIC OPERATIONS: Question 7
Simplify
75 - 27 + 48
= 63
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Question 7
Simplify
1. Find the largest “perfect square”
factor of each of the numbers!
75 - 27 + 48
75 - 27 + 48
= 25 x 3 -
9 x 3 + 16 x 3
= 53 - 33 + 43
2. Collect identical surds in same way
as you would letters.
= 63
Begin Solution
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5x – 3x + 4x = 6x
Comments
Learn Laws of Indices:
1. Find the largest “perfect square”
factor of each of the numbers!
= 25 x 3 -
a.b  a . b
9 x 3 + 16 x 3
= 53 - 33 + 43
2. Collect identical surds in same
way as you would letters.
e.g.
75  25.3
 25. 3
5 3
= 63
5x – 3x + 4x = 6x
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Comments
1. Find the largest “perfect square”
factor of each of the numbers!
= 25 x 3 -
9 x 3 + 16 x 3
= 53 - 33 + 43
2. Collect identical surds in same
way as you would letters.
The key to these simplification
questions is that all of the
individual 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 the
highest perfect square for a term.
= 63
5x – 3x + 4x = 6x
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ALGEBRAIC OPERATIONS: Question 7B
Simplify
80 + 45 - 180
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ALGEBRAIC OPERATIONS: Question 7B
Simplify
80 + 45 - 180
What would you like to do now?
Find the largest
“perfect square”
factor of each of
the numbers!
Collect identical
surds in same
way as you would
letters.
.
Reveal answer only
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ALGEBRAIC OPERATIONS: Question 7B
Simplify
80 + 45 - 180
= 5
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Question 7B
Simplify
1. Find the largest “perfect square”
factor of each of the numbers!
80 + 45 - 180
= 16 x 5 + 9 x 5 - 36 x 5
80 + 45 - 180
= 45 + 35 - 65
2. Collect identical surds in same way
as you would letters.
= 5
4x + 3x – 6x = x
Begin Solution
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Comments
Learn Laws of Indices:
1. Find the largest “perfect square”
factor of each of the numbers!
a.b  a . b
80 + 45 - 180
= 16 x 5 + 9 x 5 - 36 x 5
= 45 + 35 - 65
2. Collect identical surds in same
way as you would letters.
e.g.
80  16.5
 16. 5
4 5
= 5
4x + 3x – 6x = x
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Comments
1. Find the largest “perfect square”
factor of each of the numbers!
80 + 45 - 180
= 16 x 5 + 9 x 5 - 36 x 5
= 45 + 35 - 65
The key to these simplification
questions is that all of the
individual 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 the
highest perfect square for a term.
2. Collect identical surds in same
way as you would letters.
= 5
4x + 3x – 6x = x
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ALGEBRAIC OPERATIONS: Question 7C
Simplify
75 - 27 + 48
300 - 12
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ALGEBRAIC OPERATIONS: Question 7C
Simplify
75 - 27 + 48
300 - 12
What would you like to do now?
Deal with top and
For each,
bottom
linesfind the
Now bring
both
largest
“perfect
separately.
parts factor
together
square”
of
again.
each Cancel
of the out
where
the same
numbers!
basic surd is in
evidence on both
top and bottom.
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ALGEBRAIC OPERATIONS: Question 7C
Simplify
75 - 27 + 48
300 - 12
=
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3/
4
Question 7C
Simplify
1. Deal with top and bottom lines
separately. For each, find the
largest “perfect square” factor of
each of the numbers!
Top Line:
75 - 27 + 48
300 - 12
75 - 27 + 48
= 25 x 3 - 9 x 3 + 16 x 3
= 53 - 33 + 43
= 63
Begin Solution
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5x - 3x + 4x = 6x
Question 7C
Simplify
1. Deal with top and bottom lines
seperately. For each, find the
largest “perfect square” factor of
each of the numbers!
Bottom Line:
75 - 27 + 48
300 - 12
300 - 12
= 100 x 3 - 4 x 3
= 103 - 23
= 83
Begin Solution
Continue Solution
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10x - 2x = 8x
Question 7C
Simplify
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
75 - 27 + 48
300 - 12
3
=
4
Begin Solution
Continue Solution
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=
63
83
3/
4
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Comments
Learn Laws of Indices:
1. Deal with top and bottom lines
seperately. For each, find the
largest “perfect square” factor of
each of the numbers!
Top Line:
a.b  a . b
e.g.
75 - 27 + 48
= 25 x 3 - 9 x 3 + 16 x 3
= 53 - 33 + 43
80  16.5
 16. 5
4 5
= 63
5x - 3x + 4x = 6x
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Comments
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
=
63
83
In questions involving
fractions always treat each
line separately.
Reduce each line to its
simplest form before dividing.
The key to these simplification
questions is that all of the
individual terms can be reduced
to a multiple of the same basic
surd. In fractions these will then
often cancel.
Remember to cancel numbers too!!
=
3/
4
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ALGEBRAIC OPERATIONS: Question 8
Express
12
6
with a rational denominator.
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ALGEBRAIC OPERATIONS: Question 8
Express
12
6
with a rational denominator.
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To change the look
of a fraction but
not the value
multiply top &
bottom by the
same amount ie
6.
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a a a
ALGEBRAIC OPERATIONS: Question 8
Express
12
6
with a rational denominator.
= 26
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Question 8
Express
12
6
with a rational
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
=
6
x 6
6
2
126
=
6
1
= 26
Begin Solution
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a a a
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
6.
12
6
12
=
6
x 6
6
Rationalising the denominator:
1
a
Required to remove
from the
denominator.
Multiply top and bottom by
126
=
6
1
a
a
.

a
a a
= 26
a a a
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a
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
6.
12
6
12
=
6
x 6
6
Rationalising the denominator:
e.g.
1
1
5
5

.

5
5
5 5
126
=
6
= 26
a a a
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ALGEBRAIC OPERATIONS: Question 8B
Simplify
84
3
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ALGEBRAIC OPERATIONS: Question 8B
Simplify
84
3
What would you like to do now?
Find the largest
“perfect square”
factor of each of
the numbers!
Collect identical
surds in same
way as you would
letters.
.
Reveal answer only
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ALGEBRAIC OPERATIONS: Question 8B
Simplify
84
3
= 27
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Question 8B
1. Re-write expression using laws of
surds
Simplify 84
3
84
3
=

84
3
a.b  a . b
= 28
= 4 x 7
a
a

b
b
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= 27
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Comments
1. Re-write expression using laws of
surds
84
3
=

84
3
= 28
= 4 x 7
= 27
a
a

b
b
a.b  a . b
Learn Laws of Surds:
a
a

b
b
e.g.
24
24

 42
6
6
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Comments
1. Re-write expression using laws of
surds
84
3
=

84
3
= 28
= 4 x 7
a.b  a . b
Learn Laws of Surds:
a.b  a . b
e.g.
75  25.3  25. 3  5 3
= 27
a
a

b
b
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ALGEBRAIC OPERATIONS: Question 8C
Find the value of tanx° giving your answer as a fraction
with a rational denominator.
3m
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x°
62m
ALGEBRAIC OPERATIONS: Question 8C
Find the value of tanx° giving your answer as a fraction
with a rational denominator.
When dealing
with
To change
the look
of a
opp
0
fractions
tan
x  butcheck
fraction
not the
adj have
thatmultiply
you
value
top &
simplified
as
bottom
by as
thefar
same
possible.
amount
ie 2.
3m
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x°
62m
ALGEBRAIC OPERATIONS: Question 8C
Find the value of tanx° giving your answer as a fraction
with a rational denominator.
3m
2
=
4
x°
62m
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Question 8C
Find the value of tanx°
1. To change the look of a fraction but
not the value multiply top & bottom
by the same amount ie 2.
giving your answer as
tanx° =
a fraction with a
rational denominator.
3m
x°
62m
Begin Solution
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tan x0 
3
opp
adj
62
3
2
x
=
62
2
32
=
6x2
1
32
=
4 12
Simplify!!!
2
=
4
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Comments
1. To change the look of a fraction
but not the value multiply top &
bottom by the same amount ie
2.
tanx° =
3
62
3
2
x
=
62
2
32
=
6x2
32
=
12
2
=
4
Learn Laws of Surds:
Rationalising the denominator:
1
a
Required to remove
from the
denominator.
Multiply top and bottom by
1
a
a
.

a
a a
Next Comment
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a
Comments
1. To change the look of a fraction
but not the value multiply top &
bottom by the same amount ie
2.
tanx° =
62
3
2
x
=
62
2
32
=
6x2
2
=
4
e.g.
2
2
3 2 3

.

3
3
3 3
3
32
=
12
Learn Laws of Surds:
In questions involving fractions
make sure you have simplified
as far as possible.
Simplify!!!
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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 3 :
Further Trig
Please choose a question to attempt from the following:
1
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2
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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.
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y = asinbx°
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1200
2400
3600
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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.
0
b = no. of
waves
in 360
a = amplitude
=½
vertical
extent.
y = asinbx°
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1200
2400
3600
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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.
a=2
b=3
y = asinbx°
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1200
2400
3600
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Question 1
1. a = amplitude = ½ vertical extent.
y = asinbx°
1200
2400
3600
max /min = ±2 so
a=2
2. b = no. of waves in 3600
3 complete waves from 0 to 360
Write down the values of
a and b.
Begin Solution
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so b = 3
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Comments
Learn basic trig graphs:
1. a = amplitude = ½ vertical extent.
i.e.
max /min = ±2 so
a=2
y = sinx˚
y
1
2. b = no. of waves in 3600
x
360
3 complete waves from 0 to 360
so b = 3
-1
Max. = 1
Min. = -1
One cycle in 360˚
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Comments
Learn basic trig graphs:
1. a = amplitude = ½ vertical extent.
i.e.
max /min = ±2 so
a=2
y = sinx˚
y
1
2. b = no. of waves in 3600
x
360
One cycle
3 complete waves from 0 to 360
so b = 3
-1
y = asinx˚
Stretch
factor
y = sinbx˚
Number of
cycles in 360˚
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Further Trig : Question 1B
The graph below shows a curve with equation in the form y = acosbx° .
Write down the values of a and b.
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y = acosbx°
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900
1800
2700
3600
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Further Trig Menu
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Further Trig : Question 1B
The graph below shows a curve with equation in the form y = acosbx° .
Write down the values of a and b.
0
b = no. of
waves
in 360
a = amplitude
=½
vertical
extent.
y = acosbx°
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900
1800
2700
3600
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Further Trig Menu
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Further Trig : Question 1B
The graph below shows a curve with equation in the form y = acosbx° .
Write down the values of a and b.
a=5
b=2
y = acosbx°
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900
1800
2700
3600
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Further Trig Menu
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Question 1B
1. a = amplitude = ½ vertical extent.
y = acosbx°
900
1800
2700
3600
max /min = ±5 so
a=5
2. b = no. of waves in 3600
2 complete waves from 0 to 360
Write down the values of
a and b.
Begin Solution
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so b = 2
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Comments
Learn basic trig graphs:
1. a = amplitude = ½ vertical extent.
i.e.
max /min = ±5 so
a=5
y = cosx˚
y
1
2. b = no. of waves in 3600
2 complete waves from 0 to 360
so b = 2
x
360
-1
Max. = 1
Min. = -1
One cycle in 360˚
Next Comment
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Comments
Learn basic trig graphs:
1. a = amplitude = ½ vertical extent.
i.e.
max /min = ±2 so
a=2
y = cosx˚
y
1
2. b = no. of waves in 3600
x
360
One cycle
3 complete waves from 0 to 360
so b = 3
-1
y = acosx˚
Stretch
factor
y = cosbx˚
Number of
cycles in 360˚
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Further Trig : Question 2
Solve
5sinx° - 1 = 1
where 0<x<360
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Further Trig : Question 2
Solve
5sinx° - 1 = 1
where 0<x<360
What would you like to do now?
Always re-arrange
given equation to:
sinUse
x = CAST
diagram
cosx
= to decide
Remember
that
which
tananswers
x=quadrants
to trig
angles lie.come
equations
in pairs.
Reveal answer only
Go to full solution
Go to Comments
Further Trig Menu
EXIT
Further Trig : Question 2
Solve
5sinx° - 1 = 1
where 0<x<360
x = 23.6°
x = 156.4°
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Try another like this
Go to full solution
Go to Comments
EXIT
Further Trig Menu
Question 2
Solve
5sinx° - 1 = 1
1. Always re-arrange to sinx0 = …...
5sinx° - 1 = 1
5sinx° = 2
sinx = 0.4
where 0<x<360
2. Use the CAST diagram to decide
which quadrant angles lie in.
180 - 
sin
Begin Solution
Continue Solution
Comments
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Q1 or Q2
180 + 
tan

all
360 - 
cos
Where is
sin
positive?
Question 2
Solve
5sinx° - 1 = 1
1. Always re-arrange to sinx0 = …...
5sinx° - 1 = 1
5sinx° = 2
sinx = 0.4
where 0<x<360
Q1 or Q2
3. Calculate angles remembering that
answers to trig equations come in
pairs.
sin-1 0.4 = 23.6
Q1: angle = 23.6°
Begin Solution
Try another like this
Comments
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Q2: angle = 180° - 23.6°
= 156.4°
What would you like to do now?
Comments
We are finding the two angles at
which the sine curve reaches
a height of 0.4
1. Always re-arrange to sinx0 = …...
5sinx° - 1 = 1
5sinx° = 2
sinx = 0.4
1
Q1 or Q2
3. Calculate angles remembering
that answers to trig equations
come in pairs.
sin-1 0.4 = 23.6
y
0.5
0.4
x
23.6
90
156.4
180
270
-0.5
-1
Q1: angle = 23.6°
Q2: angle = 180° - 23.6°
Try another
Menu
= 156.4°
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360
Further Trig : Question 2B
Solve
3tanx° + 7 = 2
where 0<x<360
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Get hint
Reveal answer only
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Further Trig Menu
EXIT
Further Trig : Question 2B
Solve
3tanx° + 7 = 2
where 0<x<360
What would you like to do now?
Always re-arrange
given equation to:
sinUse
x = CAST
diagram
cosx
= to decide
Remember
that
which
tananswers
x=quadrants
to trig
angles lie.come
equations
in pairs.
Reveal answer only
Go to full solution
Go to Comments
Further Trig Menu
EXIT
Further Trig : Question 2B
Solve
3tanx° + 7 = 2
where 0<x<360
x = 121.0°
x = 301.0°
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Further Trig Menu
EXIT
Question 2B
Solve
1. Always re-arrange to tanx0 = …...
3tanx° + 7 = 2
3tanx° = -5
tanx =
3tanx° + 7 = 2
where 0<x<360
Continue Solution
Comments
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3
Q2 or Q4
2. Use the CAST diagram to decide
which quadrant angles lie in.
180 - 
sin
Begin Solution
-5/
180 + 
tan

all
360 - 
cos
Where is
tan
negative?
Question 2B
Solve
3tanx° + 7 = 2
where 0<x<360
What would you like to do now?
Begin Solution
1. Always re-arrange to tanx0 = …...
3tanx° + 7 = 2
3tanx° = -5
tanx =
-5/
3
Q2 or Q4
3. Calculate angles remembering that
answers to trig equations come in
pairs.
tan-1 (53) = 59.0
Q2: angle = 180° - 59.0°
= 121.0°
Continue Solution
Comments
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Q4: angle = 360° - 59.0°
= 301.0°
Comments
We are finding the two angles
at which the tan graph reaches
5
a height of
1. Always re-arrange to tanx0 = …...

3tanx° + 7 = 2
3tanx° = -5
tanx =
-5/
3
3 y
2
3
1
3. Calculate angles remembering that
answers to trig equations come in
pairs.
tan-1 (53) = 59.0
Q2: angle = 180° - 59.0°
= 121.0°
Q4: angle = 360° - 59.0°
= 301.0°
5

3
301.0
121.0
90
180
270
-1
-2
-3
Next Comment
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x
360
Comments
1. Always re-arrange to tanx0 = …...
Note:
Never put a negative
value into the
inverse trig function
3tanx° + 7 = 2
3tanx° = -5
tanx =
-5/
3
3. Calculate angles remembering that
answers to trig equations come in
pairs.
tan-1
(53) = 59.0
Q2: angle = 180° - 59.0°
= 121.0°
Q4: angle = 360° - 59.0°
= 301.0°
DO:
5
x  tan  
 3
1
then CAST diagram
Next Comment
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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 3 :
Quadratic
Functions
Please choose a question to attempt from the following:
1
EXIT
2
3
Back to
Unit 3 Menu
4
Quadratic Functions: Question 1
Solve the equation x2 – 6x + 7 = 0 giving your
answers to 2 decimal places.
What would you like to do now?
Get hint
Reveal answer only
Go to full solution
Go to Comments
Go to Quadratics Menu
EXIT
Quadratic Functions: Question 1
Solve the equation x2 – 6x + 7 = 0 giving your
answers to 2 decimal places.
Write down
values of a,
What would you like to do now?
b & c.b2 – 4ac .
Evaluate
Remember to
Now use
round.
Reveal answer only
quadratic
formula,
Go to full solution
rememberin
g that you
Go to Comments
should get
two
Go to Quadratics Menu
answers.
EXIT
Quadratic Functions: Question 1
Solve the equation x2 – 6x + 7 = 0 giving your
answers to 2 decimal places.
x = 4.41 or 1.59
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Try another like this
Go to full solution
Go to Comments
Go to Quadratics Menu
EXIT
Question 1
Solve the equation
1. Write down values of a, b & c.
x2 – 6x + 7 = 0
x2 – 6x + 7 = 0
giving your answers
to 2 decimal places.
a = 1 b = (-6) c = 7
2. Evaluate b2 – 4ac .
b2 – 4ac = (-6)2 – (4x1x7)
= 36 – 28
=8
Begin Solution
Continue Solution
Comments
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Question 1
Solve the equation
1. Write down values of a, b & c.
x2 – 6x + 7 = 0
x2 – 6x + 7 = 0
giving your answers
to 2 decimal places.
a = 1 b = -6 c = 7
3. Now use quadratic formula.
x = -b ± (b2 – 4ac )
2a
Begin Solution
Continue Solution
Comments
Menu
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x = 6 ± 8
2
4. Rewrite with brackets and now use
calculator.
= (6 + 8)  2 or (6 - 8)  2
Question 1
Solve the equation
1. Write down values of a, b & c.
x2 – 6x + 7 = 0
x2 – 6x + 7 = 0
giving your answers
to 2 decimal places.
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..
Begin Solution
Try another like this
5. Remember to round.
x = 4.41 or 1.59
Comments
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Back to Home
What would you like to do now?
Comments
1. Write down values of a, b & c.
x2 – 6x + 7 = 0
a = 1 b = -6 c = 7
3. Now use quadratic formula.
x = -b ± (b2 – 4ac )
2a
x = 6 ± 8
2
4. Rewrite with brackets and now use
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!!
Next Comment
Menu
Back to Home
Comments
1. Write down values of a, b & c.
x2 – 6x + 7 = 0
Refer to the Formula Sheet:
ax2 + bx + c = 0
Quadratic Formula:
a = 1 b = -6 c = 7
3. Now use quadratic formula.
x = -b ± (b2 – 4ac )
2a
x = -b ± (b2 – 4ac )
2a
x = 6 ± 8
2
4. Rewrite with brackets and now use
calculator.
= (6 + 8)  2 or (6 - 8)  2
Next Comment
Menu
Back to Home
Comments
1. Write down values of a, b & c.
x2 – 6x + 7 = 0
Take care when allocating a
value to a, b and c.
ax2 + bx + c = 0
a = 1 b = -6 c = 7
3. Now use quadratic formula.
x = -b ± (b2 – 4ac )
2a
x = 6 ± 8
2
4. Rewrite with brackets and now use
calculator.
= (6 + 8)  2 or (6 - 8)  2
1x2 - 6x + 7 = 0
a =1
b = (-6)
c=7
Next Comment
Menu
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Comments
1. Write down values of a, b & c.
x2
– 6x + 7 = 0
a = 1 b = -6 c = 7
Note:
Finding b2 – 4ac
working.
first eases
Bracket all negative numbers.
Watch out for double negative!
2. Evaluate b2 – 4ac .
b2 – 4ac = (-6)2 – (4x1x7)
= 36 – 28
=8
Try another
Menu
Back to Home
Quadratic Functions: Question 1B
Solve the equation 3x2 + 5x - 1 = 0 giving your
answers to 2 significant figures.
What would you like to do now?
Get hint
Reveal answer only
Go to full solution
Go to Comments
Go to Quadratics Menu
EXIT
Quadratic Functions: Question 1B
Solve the equation 3x2 + 5x - 1 = 0 giving your
answers to 2 significant figures.
Write down
values of a,
b & c.b2 – 4ac .
Evaluate
Remember to
Now use
round.
quadratic
What would you like to do now?
formula,
rememberin
Reveal answer only
g that you
should get
Go to full solution
two
answers.
Go to Comments
Go to Quadratics Menu
EXIT
Quadratic Functions: Question 1B
Solve the equation 3x2 + 5x - 1 = 0 giving your
answers to 2 significant figures.
x = 0.18 or -1.8
What would you like to do now?
Go to full solution
Go to Comments
Go to Quadratics Menu
EXIT
Question 1B
Solve the equation
3x2 + 5x - 1 = 0
1. Write down values of a, b & c.
3x2 + 5x – 1 = 0
a = 3 b = 5 c = (-1)
2. Evaluate b2 – 4ac .
giving your answers
to 2 significant figures.
b2 – 4ac = (5)2 – (4 x 3 x -1)
= 25 – (- 12)
= 37
Begin Solution
Continue Solution
Comments
Quadratics Menu
Back to Home
Question 1B
Solve the equation
3x2 + 5x - 1 = 0
1. Write down values of a, b & c.
3x2 + 5x – 1 = 0
a = 3 b = 5 c = (-1)
3. Now use quadratic formula.
giving your answers
to 2 significant figures.
Begin Solution
Continue Solution
Comments
Quadratics Menu
Back to Home
x = -b ± (b2 – 4ac )
2a
x = -5 ± 37
2
4. Rewrite with brackets and now use
calculator.
= (-5 + 37)  2 or (-5 - 37)  2
Question 1B
Solve the equation
3x2 + 5x - 1 = 0
giving your answers
to 2 significant figures.
1. Write down values of a, b & c.
3x2 + 5x – 1 = 0
a = 3 b = 5 c = (-1)
4. Rewrite with brackets and now use
calculator.
= (-5 + 37)  2 or (-5 - 37)  2
= 0.180… or -1.847..
Begin Solution
Continue Solution
5. Remember to round.
x = 0.18 or -1.8
Comments
Quadratics Menu
Back to Home
What would you like to do now?
Comments
1. Write down values of a, b & c.
3x2 + 5x – 1 = 0
a=3 b=5
c = (-1)
3. Now use quadratic formula.
x = -b ± (b2 – 4ac )
2a
x = -5 ± 37
2
4. Rewrite with brackets and now
use calculator.
= (-5 + 37)  2 or (-5 - 37)  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!!
Next Comment
Quadratics Menu
Back to Home
Comments
1. Write down values of a, b & c.
3x2
+ 5x – 1 = 0
Refer to the Formula Sheet:
ax2 + bx + c = 0
Quadratic Formula:
a=3 b=5
c = (-1)
3. Now use quadratic formula.
x = -b ± (b2 – 4ac )
2a
x = -b ± (b2 – 4ac )
2a
x = -5 ± 37
2
4. Rewrite with brackets and now use
calculator.
= (-5 + 37)  2 or (-5 - 37)  2
Next Comment
Quadratics Menu
Back to Home
Comments
1. Write down values of a, b & c.
3x2 + 5x – 1 = 0
Take care when allocating a
value to a, b and c.
ax2 + bx + c = 0
a=3 b=5
c = (-1)
3. Now use quadratic formula.
x = -b ± (b2 – 4ac )
2a
3x2 + 5x - 1 = 0
a =3
b=5
c = -1
x = -5 ± 37
2
4. Rewrite with brackets and now use
calculator.
= (-5 + 37)  2 or (-5 - 37)  2
Next Comment
Quadratics Menu
Back to Home
Comments
Note:
1. Write down values of a, b & c.
3x2 + 5x – 1 = 0
a = 3 b = 5 c = (-1)
Finding b2 – 4ac
working.
first eases
Bracket all negative numbers.
Watch out for double negative!
2. Evaluate b2 – 4ac .
b2 – 4ac = (5)2 – (4 x 3 x -1)
= 25 – (- 12)
= 37
Next Comment
Quadratics Menu
Back to Home
Quadratic Functions: Question 2
Solve the equation 3x2 = x + 1 giving your
answers to 2 decimal places.
What would you like to do now?
Get hint
Reveal answer only
Go to full solution
Go to Comments
Go to Quadratics Menu
EXIT
Quadratic Functions: Question 2
Solve the equation 3x2 = x + 1 giving your
answers to 2 decimal places.
Write down
values of a,
b Re-write
& c.b2 – 4ac .
Evaluate
quadratic: to
Remember
Now
make
it use
round.
quadratic
What would you like to do now?
equal
to
formula,
zero before
rememberin
solving.
Reveal answer only
g that you
should get
Go to full solution
two
answers.
Go to Comments
Go to Quadratics Menu
EXIT
Quadratic Functions: Question 2
Solve the equation 3x2 = x + 1 giving your
answers to 2 decimal places.
x = 0.77 or -0.43
What would you like to do now?
Try another like this
Go to full solution
Go to Comments
Go to Quadratics Menu
EXIT
Question 2
1. Rearrange in quadratic form.
3x2 = x + 1
Solve the equation
3x2 – x – 1 = 0
3x2 = x + 1
giving your answers
a = 3 b = (-1) c = (-1)
to 2 decimal places.
2. Write down values of a, b & c.
3. Evaluate b2 – 4ac .
b2 – 4ac = (-1)2 – (4 x 3 x -1)
Begin Solution
= 1 – (-12)
Continue Solution
Comments
Quadratics Menu
Back to Home
= 13
Question 2
Solve the equation
3x2 = x + 1
2. Write down values of a, b & c.
3x2 –x – 1 = 0
a = 3 b = (-1) c = (-1)
giving your answers
to 2 decimal places.
3. Now use quadratic formula.
x = -b ± (b2 – 4ac )
2a
Begin Solution
Continue Solution
Comments
Quadratics Menu
Back to Home
x = 1 ± 13
2
4. Rewrite with brackets and now use
calculator.
= (1 + 13)  2 or (1 - 13)  2
Question 2
Solve the equation
3x2 = x + 1
1. Write down values of a, b & c.
3x2 – x – 1 = 0
a = 3 b = (-1) c = (-1)
giving your answers
to 2 decimal places.
4. Rewrite with brackets and now use
calculator.
= (1 + 13)  2 or (1 - 13)  2
= 0.767… or -0.434..
Begin Solution
Try another like this
5. Remember to round.
x = 0.77 or -0.43
Comments
Quadratics Menu
Back to Home
What would you like to do now?
Comments
1. Rearrange in quadratic form.
3x2 = x + 1
3x2 – x – 1 = 0
For any question involving a
quadratic in which you are
asked to give your answer to
a given number of decimal
places
OR
a = 3 b = (-1) c = (-1)
2. Write down values of a, b & c.
3. Evaluate
b2
– 4ac .
a given number of significant
figures:
THINK QUADRATIC
FORMULA!!
b2 – 4ac = (-1)2 – (4 x 3 x -1)
= 1 – (-12)
Next Comment
= 13
Quadratics Menu
Back to Home
Comments
1. Rearrange in quadratic form.
3x2 = x + 1
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)
2. Write down values of a, b & c.
3. Evaluate b2 – 4ac .
b2 – 4ac = (-1)2 – (4 x 3 x -1)
= 1 – (-12)
Next Comment
= 13
Quadratics Menu
Back to Home
Comments
2. Write down values of a, b & c.
3x2 –x – 1 = 0
Refer to the Formula Sheet:
ax2 + bx + c = 0
Quadratic Formula:
a = 3 b = (-1) c = (-1)
3. Now use quadratic formula.
x = -b ± (b2 – 4ac )
2a
x = -b ± (b2 – 4ac )
2a
x = 1 ± 13
2
4. Rewrite with brackets and now use
calculator.
= (1 + 13)  2 or (1 - 13)  2
Next Comment
Quadratics Menu
Back to Home
Comments
2. Write down values of a, b & c.
3x2 –x – 1 = 0
Take care when allocating a
value to a, b and c.
ax2 + bx + c = 0
a = 3 b = (-1) c = (-1)
3. Now use quadratic formula.
x = -b ± (b2 – 4ac )
2a
x = 1 ± 13
2
4. Rewrite with brackets and now use
calculator.
= (1 + 13)  2 or (1 - 13)  2
3x2 - 1x - 1 = 0
a =3
b = (-1)
c = (-1)
Next Comment
Quadratics Menu
Back to Home
Comments
1. Rearrange in quadratic form.
3x2
=x+1
3x2 – x – 1 = 0
Note:
Finding b2 – 4ac
working.
first eases
Bracket all negative numbers.
Watch out for double negative!
a = 3 b = (-1) c = (-1)
2. Write down values of a, b & c.
3. Evaluate b2 – 4ac .
b2 – 4ac = (-1)2 – (4 x 3 x -1)
= 1 – (-12)
= 13
Try another like this
Quadratics Menu
Back to Home
Quadratic Functions: Question 2B
Solve the equation
2x(x + 2) = 2 - x
giving your answers to 2 significant figures.
What would you like to do now?
Get hint
Reveal answer only
Go to full solution
Go to Comments
Go to Quadratics Menu
EXIT
Quadratic Functions: Question 2B
Solve the equation
2x(x + 2) = 2 - x
giving your answers to 2 significant figures.
Write
Evaluate
b2 down
– 4ac .
Remember
to
values of a,
Now use
Re-write
round.
b & c.
quadratic
What would you like to do now? quadratic:
formula,
get rid
of
rememberin
brackets
&
Reveal answer only
g that
make
it you
should
equal
to get
Go to full solution
two
zero before
answers.
solving.
Go to Comments
Go to Quadratics Menu
EXIT
Quadratic Functions: Question 2B
Solve the equation
2x(x + 2) = 2 - x
giving your answers to 2 significant figures.
x = 0.35 or -2.9
What would you like to do now?
Go to full solution
Go to Comments
Go to Quadratics Menu
EXIT
Question 2B
Solve the equation
2x(x + 2) = 2 - x
1. Rearrange in quadratic form.
2x(x + 2) = 2 – x
2x2 + 4x – 2 + x= 0
2x2 + 5x – 2 = 0
giving your answers
to 2 significant figures.
a = 2 b = 5 c = (-2)
2. Write down values of a, b & c.
3. Evaluate b2 – 4ac .
Begin Solution
Continue Solution
b2 – 4ac = (5)2 – (4 x 2 x -2)
= 25 – (-16)
Comments
= 41
Quadratics Menu
Back to Home
Question 2B
Solve the equation
2x(x + 2) = 2 - x
giving your answers
to 2 significant figures.
1. Rearrange in quadratic form.
2x2 + 5x – 2 = 0
a = 2 b = 5 c = (-2)
3. Now use quadratic formula.
x = -b ± (b2 – 4ac )
2a
Begin Solution
Continue Solution
Comments
Quadratics Menu
Back to Home
x = -5 ± 41
2
4. Rewrite with brackets and now use
calculator.
= (-5 + 41)  2 or (-5 - 41)  2
Question 2B
Solve the equation
1. Rearrange in quadratic form.
2x2 + 5x – 2 = 0
2x(x + 2) = 2 - x
giving your answers
to 2 significant figures.
a = 2 b = 5 c = (-2)
4. Rewrite with brackets and now use
calculator.
= (-5 + 41)  2 or (-5 - 41)  2
= 0.350… or -2.850..
Begin Solution
Continue Solution
5. Remember to round.
x = 0.35 or -2.9
Comments
Quadratics Menu
Back to Home
What would you like to do now?
Comments
1. Rearrange in quadratic form.
2x(x + 2) = 2 – x
2x2 + 4x – 2 + x= 0
2x2 + 5x – 2 = 0
a = 2 b = 5 c = (-2)
2. Write down values of a, b & c.
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!!
3. Evaluate b2 – 4ac .
b2 – 4ac = (5)2 – (4 x 2 x -2)
= 25 – (-16)
= 41
Next Comment
Quadratics Menu
Back to Home
Comments
1. Rearrange in quadratic form.
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.
2x2 + 5x – 2 = 0
a = 2 b = 5 c = (-2)
2. Write down values of a, b & c.
3. Evaluate b2 – 4ac .
b2 – 4ac = (5)2 – (4 x 2 x -2)
= 25 – (-16)
= 41
Next Comment
Quadratics Menu
Back to Home
Comments
1. Rearrange in quadratic form.
2x2 + 5x – 2 = 0
Refer to the Formula Sheet:
ax2 + bx + c = 0
Quadratic Formula:
a=2 b=5
c = (-2)
3. Now use quadratic formula.
x = -b ± (b2 – 4ac )
2a
x = -b ± (b2 – 4ac )
2a
x = -5 ± 41
2
4. Rewrite with brackets and now
use calculator.
= (-5 + 41)  2 or (-5 - 41)  2
Next Comment
Quadratics Menu
Back to Home
Comments
1. Rearrange in quadratic form.
2x2 + 5x – 2 = 0
Take care when allocating a
value to a, b and c.
ax2 + bx + c = 0
a=2 b=5
c = (-2)
2x2 + 5x - 2 = 0
3. Now use quadratic formula.
x = -b ± (b2 – 4ac )
2a
a =2
b=5
c = -2
x = -5 ± 41
2
4. Rewrite with brackets and now
use calculator.
= (-5 + 41)  2 or (-5 - 41)  2
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Comments
1. Rearrange in quadratic form.
2x(x + 2) = 2 – x
2x2 + 4x – 2 + x= 0
2x2 + 5x – 2 = 0
Note:
Finding b2 – 4ac
working.
first eases
Bracket all negative numbers.
Watch out for double negative!
a = 2 b = 5 c = (-2)
2. Write down values of a, b & c.
3. Evaluate b2 – 4ac .
b2 – 4ac = (5)2 – (4 x 2 x -2)
= 25 – (-16)
= 41
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Quadratic Functions: Question 3
The diagram below shows an L-shaped plot of land with dimensions
as given.
xm
(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
xm
(x+3)
m
EXIT
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Quadratic Functions: Question 3
The diagram below shows an L-shaped plot of land with dimensions
as given.
xm
(a) Show that the total area is given by the
expression
x2 + 7x m2.
(x+4) m
EXIT
(b) Hence find the value of x when this area is
60 m2.
Find the
In of
(b)each
make
Make
area
Add
these
to
findit
xm
expression
equal to
rectangle..
total area.
from (a)zero,
=
(x+3)
60.
factorise &
m
solve.
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Reveal answer
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Quadratic Functions: Question 3
The diagram below shows an L-shaped plot of land with dimensions
as given.
xm
(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 = 5m
xm
(x+3)
m
What would you like to do now?
EXIT
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Question 3
1. Find the area of each rectangle
separately and add together to get
total area.
xm
(x+4)
m
(a)
3m
1
Area rectangle
2
1
xm
(x+3)
m
(a) Show that the total area
is given by the
expression x2 + 7x m2.
= x(x+4)
= x2 + 4x m2
Length rectangle 2 = (x + 3) – x
= 3m
Area rectangle
2
= 3x m2
Begin Solution
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Hence total area = (x2 + 4x) + 3x
= x2 + 7x m2
Question 3
2. Use the expression for area given in
part (a) and make it equal to 60.
xm
(b)
(x+4)
m
If total area = 60
And total area = x2 + 7x
1
then
2
xm
(x+3)
m
(b) Hence find the value of x
m2.
when this area is 60
What would you like to do now?
x2 + 7x = 60
3. Make the equation equal to zero,
factorise and solve.
x2 + 7x - 60 = 0
(x + 12)(x – 5) = 0
So (x + 12) = 0 or (x – 5) = 0
Begin Solution
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Comments
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ie
x = -12 or
x = 5
Length must be 5m as negative
value not valid.
Comments
2. Use the expression for area given
in part (a) and make it equal to
60.
(b) If total area = 60
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
x = -12 or
x = 5
Length must be 5m as negative
value not valid.
Refer to the Formula Sheet:
ax2 + bx + c = 0
3. Make the equation equal to zero,
factorise and solve.
ie
The quadratic equation can also
be solved by applying
the quadratic formula:
Quadratic Formula:
x = -b ± (b2 – 4ac )
2a
Finding b2 – 4ac
first
eases working.
Bracket all negative numbers.
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Comments
2. Use the expression for area given
in part (a) and make it equal to
60.
(b) If total area = 60
And total area = x2 + 7x
then
x2 + 7x = 60
Take care when allocating a
value to a, b and c.
ax2 + bx + c = 0
1x2 + 7x - 60 = 0
3. Make the equation equal to zero,
factorise and solve.
x2 + 7x - 60 = 0
(x + 12)(x – 5) = 0
a =1
b=7
c = (-60)
So (x + 12) = 0 or (x – 5) = 0
ie
x = -12 or
x = 5
Length must be 5m as negative
value not valid.
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Comments
2. Use the expression for area given
in part (a) and make it equal to
60.
(b) If total area = 60
And total area = x2 + 7x
then
x2 + 7x = 60
3. Make the equation equal to zero,
factorise and solve.
x2 + 7x - 60 = 0
Make sure that you
understand when a
negative answer
may not be
correct in context
given,
e.g. negative lengths
or areas.
(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.
Try another like this
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Quadratic Functions: Question 3B
The diagram below shows plans of the foundations for a house and
garage.
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.
(a) Find an expression for the
area of the garage in terms of x.
7m
garage ? m
house
(b) Hence find the value of x
xm
12m
Get hint
EXIT
to meet this requirement.
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Quadratic Functions: Question 3B
The diagram below shows plans of the foundations for a house and
garage.
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.
Area =
(a) Find an expression
length x for the
area of thewidth.
garage
in terms of x.
Use
garage ? m
information
7m
house
Use Make it
find
Find
thetonumerical
expression
equal
to of x
(b) Hence
find
the
value
length
of
value
of
the
from (a) to
zero,
x
m
garage.
required
area
for
to meetform
thisfactorise
12m
arequirement.
&
the
garage.
quadratic
solve.
equation.
What would you like to do now?
EXIT
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Quadratic Functions: Question 3B
The diagram below shows plans of the foundations for a house and
garage.
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.
(a) Find an expression for the
area of the garage in terms of x.
7m
house
garage ? m
= x2 + 4x m2
(b) Hence find the value of x
12m
xm
to meet this requirement.
x = 3
What would you like to do now?
Go to Comments
EXIT
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Question 3B
7m
house
1. Area = length x width. Width is x but
still need to find length.
garage?
m
x
12m
m
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.
Begin Solution
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(a)
Length of garage = (x+4) m
So area = x(x+4)
= x2 + 4x m2
Question 3B
7m
house
garage?
m
2. Area = length x width. Find the
required area and use your answer
to (a) to establish quadratic
equation.
(b) Area house = 12 x 7 = 84m2
x
Area garage = 25% of 84m2
m
the length of the garage should
= 21m2
be 4m longer than its width and
So
x2 + 4x = 21
its floor area should be 25%
3. Make the equation equal to zero,
of the floor area of the house.
factorise and solve.
12m
x2 + 4x - 21 = 0
Begin Solution
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(x + 7)(x – 3) = 0
(x + 7) = 0 or (x – 3) = 0
Question 3B
7m
house
(b)
garage?
m
x
m
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.
12m
3. Make the equation equal to zero,
factorise and solve.
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.
Begin Solution
Continue Solution
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What would you like to do now?
Comments
2. Area = length x width. Find the
required area and use your
answer to (a) to establish
quadratic equation.
Area house = 12 x 7 = 84m2
Area garage = 25% of 84m2
So
= 21m2
x2 + 4x = 21
3. Make the equation equal to zero,
factorise and solve.
x2 + 4x - 21 = 0
(x + 7)(x – 3) = 0
(x + 7) = 0 or (x – 3) = 0
The quadratic equation can also
be solved by applying
the quadratic formula:
Refer to the Formula Sheet:
ax2 + bx + c = 0
Quadratic Formula:
x = -b ± (b2 – 4ac )
2a
Finding b2 – 4ac
first
eases working.
Bracket all negative numbers.
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Comments
2. Area = length x width. Find the
required area and use your
answer to (a) to establish
quadratic equation.
Area house = 12 x 7 = 84m2
Area garage = 25% of 84m2
So
= 21m2
x2 + 4x = 21
3. Make the equation equal to zero,
factorise and solve.
Take care when allocating a
value to a, b and c.
ax2 + bx + c = 0
1x2 + 4x - 21 = 0
a =1
b=4
c = (-21)
x2 + 4x - 21 = 0
(x + 7)(x – 3) = 0
(x + 7) = 0 or (x – 3) = 0
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Comments
(b)
3. Make the equation equal to zero,
factorise and solve.
x2 + 4x - 21 = 0
(x + 7)(x – 3) = 0
(x + 7) = 0 or (x – 3) = 0
ie.
x = -7
or
Make sure that you
understand when a
negative answer
may not be
correct in context
given,
e.g. negative lengths
or areas.
x = 3
Length must be 3m as
negative value not valid.
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Quadratic Functions: Question 4
The blades in electric hair-clippers have grooves in the form of
identical adjacent parabolas.
Y
C
D
X
E
F
G
The second parabola has turning point E and equation
y = (x – 5)2 – 9 .
(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.
EXIT
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Quadratic Functions: Question 4
The blades in electric hair-clippers have grooves in the form of
identical adjacent parabolas.
Y
C
D
X
E
F
G
The second parabola has turning point E and equation
y = (x – 5)2 – 9 .
In ‘completed
roots
are
use
‘completed
Use
symmetry
square’,
a of a
equidistant
(a) State the coordinates of E.
square’
parabola
to
findform
otherto
minimum
value
from equation
axis of
write
root
if
you
know
one.
when
(b) The second parabola cuts the X-axis at Coccurs
& D. D
is the point
symmetry.
bracketofisparabola.
zero.
(8,0), find the coordinates of C.
(c) Find the equation of the parabola with minimum turning point G.
What would you like to do now?
EXIT
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Quadratic Functions: Question 4
The blades in electric hair-clippers have grooves in the form of
identical adjacent parabolas.
Y
C
D
X
E
F
G
The second parabola has turning point E and equation
y = (x – 5)2 – 9 .
(a) State the coordinates of E.
E is the point (5,-9).
(b) The second parabola cuts the X-axis at C & D. D is the point (8,0),
find the coordinates of C.
C is (2,0).
(c) Find the equation of the parabola with minimum turning point G.
y = (x – 17)2 - 9
EXIT
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Question 4
Y
C
1. In ‘completed square’, a minimum
value occurs when bracket is zero.
D
X
(a)
y will have a minimum value of
E
F
G
-9 when (x – 5)2 = 0.
The second parabola has
turning point E and equation
y = (x – 5)2 – 9 .
Begin Solution
Continue Solution
Comments
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ie when x = 5
so E is the point (5,-9).
Question 4
Y
C
2. Use symmetry of a parabola to find
other root if you know one.
D
X
(b) Horizontal dist from E to D
= horizontal dist from E to C
E
F
G
= 3 units
(b) D is the point (8,0),
find the coordinates of C.
D has x-coordinate 8
so x-coordinate of C is 6 units less
E is the point (5,-9).
ie
Begin Solution
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C is (2,0).
Question 4
Y
C
D
X
3. Use symmetry of a parabola to find
other turning points if you know
one.
(c)Distance from E to G
E
F
is 12 units so G is (17,-9)
G
(c) Find the equation of the
parabola with minimum G.
E is the point (5,-9).
Begin Solution
Try another like this
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 - 9
Comments
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What would you like to do now?
Comments
Learn the following Results:
1. In ‘completed square’, a minimum
value occurs when bracket is
zero.
(a)
When a quadratic equation is in
the form y = (x – a)2 + b
the parabola has a
y will have a minimum value of
minimum T.P. at (a,b).
-9 when (x – 5)2 = 0.
ie when x = 5
so E is the point (5,-9).
When a quadratic equation is in
the form y = b - (x – a)2
the parabola has a
maximum T.P. at (a,b).
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Comments
Learn the following Results:
1. In ‘completed square’, a minimum
value occurs when bracket is
zero.
(a)
e.g.
y = (x – 5)2 - 9
Minimum T.P. at P(5,-9)
y will have a minimum value of
-9 when (x – 5)2 = 0.
ie when x = 5
so E is the point (5,-9).
P
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Comments
Learn the following Results:
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
P
D has x-coordinate 8
so y-coordinate of C is 6 units less The roots are equidistant from the
ie C is (2,0).
turning point (or axis of symmetry)
in the horizontal direction.
Try another like this
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Quadratic Functions: Question 4B
The top of an ornate fence consists of a series of congruent adjacent
parabolas.
X
U
V
W
Y
T
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.
EXIT
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Get hint
Quadratic Functions: Question 4B
The top of an ornate fence consists of a series of adjacent parabolas.
X
U
V
W
Y
T
In ‘completed
The first parabola has turning point U and equation
y‘completed
= 4of
– a(x – 3)2 .
Use
symmetry
use
square’,
parabola
toafind
square’
form to
(a) State the coordinates of U.
minimum
value
another tp if you know
write
equation
occurs
when
one.
(b) If T is the point (5,0) then find the coordinates of
of parabola.
V, the
bracket is zero.
maximum turning point on the second parabola.
(c) Find the equation of the parabola with maximum turning point
W.
What would you like to do now?
EXIT
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Quadratic Functions: Question 4B
The top of an ornate fence consists of a series of adjacent parabolas.
X
U
V
W
Y
T
The first parabola has turning point U and equation y = 4 – (x – 3)2 .
(a) State the coordinates of U.
U is the point (3,4).
(b) If T is the point (5,0) then find the coordinates of V, the
maximum turning point on the second parabola.
V is (7,4).
(c) Find the equation of the parabola with maximum turning point
W.
y = 4 – (x – 11)2
EXIT
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Question 4B
1. In ‘completed square’, a maximum
value occurs when bracket is zero.
X
U
V
(a)
W
y will have a maximum value of
T
Y
The first parabola has turning
point U and equation
y = 4 – (x – 3)2 .
Begin Solution
Continue Solution
Comments
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4 when (x – 3)2 = 0.
ie when x = 3
so U is the point (3,4).
Question 4B
2. Use symmetry of a parabola to find
another turning point if you know
one.
X
U
V
(b) Horizontal dist from U to T
W
= horizontal dist from T to V
T
(b) T is the point (5,0),
Y
find the coordinates of V.
U is the point (3,4).
Begin Solution
Continue Solution
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= 2 units
For identical parabolae,maximum
values same so y-coordinate = 4
T has x-coordinate 5
so x-coordinate of V is 2 units more
ie V is (7,4).
Question 4B
3. Use symmetry of a parabola to find
other turning points if you know
one.
X
U
V
W
(c)Distance from U to W
is 8 units so W is (11,4)
T
(c) Find the equation of the
parabola with maximum W.
U is the point (3,4).
Begin Solution
Try another like this
Y
4. If you know turning point, use
‘completed square’ form to write
equation of parabola.
Equation of this parabola
in same form as the first
ie
y = 4 - (x – 11)2
Comments
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What would you like to do now?
Comments
Learn the following Results:
1. In ‘completed square’, a maximum
value occurs when bracket is
zero.
(a)
When a quadratic equation is in
the form y = (x – a)2 + b
the parabola has a
y will have a maximum value of
minimum T.P. at (a,b).
4 when (x – 3)2 = 0.
ie when x = 3
so U is the point (3,4).
When a quadratic equation is in
the form y = b - (x – a)2
the parabola has a
maximum T.P. at (a,b).
Next Comment
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Comments
Learn the following Results:
1. In ‘completed square’, a maximum
value occurs when bracket is
zero.
e.g.
y = 4 – (x – 3)2
Maximum T.P. at P(3,4)
(a)
y will have a maximum value of
4 when (x – 3)2 = 0.
P
ie when x = 3
so U is the point (3,4).
Try another like this
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Quadratic Functions: Question 4C
The diagram below shows some identical adjacent parabolas.
Y
H
J
X
F
G
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 .
(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.
EXIT
Reveal answer
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Go to full solution
Get hint
Quadratic Functions: Question 4C
The diagram below shows some identical adjacent parabolas.
Y
H
J
X
F
G
The last parabola has a minimum turning point at G and
cuts the x-axis at H and J.
In ‘completed
2
use ‘completed
Its equation is y = (x – 12) – 4 .
If square’,
not given
aa root you
square’
must solve
the form to
minimum
value
(a) State the coordinates of G.
write
equation
quadratic.
occurs
when
(b) Find the coordinates of H and J.
bracketofisparabola.
zero.
(c) Find the equation of the parabola with minimum turning point at F.
What would you like to do now?
EXIT
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Quadratic Functions: Question 4C
The diagram below shows some identical adjacent parabolas.
Y
H
J
X
F
G
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 .
(a) State the coordinates of G.
(b) Find the coordinates of H and J.
G is the point (12,-4).
H is (10,0) and J is (14,0)
(c) Find the equation of the parabola with minimum turning point at F.
What would you like to do now?
y = (x – 8)2 - 4
Go to Comments
EXIT
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Question 4C
1. In ‘completed square’, a minimum
value occurs when bracket is zero.
Y
H
J
X
(a)
y will have a minimum value of
F
G
Its equation is y = (x – 12)2 – 4 .
-4 when (x – 12)2 = 0.
ie when x = 12
so G is the point (12,-4).
Begin Solution
Continue Solution
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Question 4C
Y
y = (x –
12)2
H
J
–4.
X
2. If you are not given one of the roots
you must solve quadratic to find
roots.
(b) At H & J
F
G
(b) Find the coordinates of
H and J.
G is the point (12,-4).
Begin Solution
Continue Solution
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(x – 12)2 – 4 = 0
so
ie
(x – 12)2 = 4
x – 12 = -2 or 2
ie
x = 10 or 14
From diagram:
H is (10,0) and J is (14,0)
Question 4C
10
H
Y
14
J
X
3. Use symmetry of a parabola to find
other turning points if you know
one.
(c)Distance from G to H
F
G
is 2 units so F is 4 units
horizontally from G.
So F is the point (8,-4).
(c) Find the equation of the
parabola with minimum F.
G is the point (12,-4).
Begin Solution
Continue Solution
4. If you know turning point, use
‘completed square’ form to write
equation of parabola.
Equation of this parabola
in same form as the last
Comments
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ie
y = (x – 8)2 - 4
What would you like to do now?
Comments
Learn the following Results:
1. In ‘completed square’, a minimum
value occurs when bracket is
zero.
When a quadratic equation is in
(a)
the parabola has a
y will have a minimum value of
minimum T.P. at (a,b).
-4 when (x – 12)2 = 0.
ie when x = 12
so G is the point (12,-4).
the form y = (x – a)2 + b
When a quadratic equation is in
the form y = b - (x – a)2
the parabola has a
maximum T.P. at (a,b).
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Learn the following Results:
1. In ‘completed square’, a minimum
value occurs when bracket is
zero.
(a)
e.g.
y = (x – 12)2 – 4
Minimum T.P. at P(12,-4)
y will have a minimum value of
-4 when (x – 12)2 = 0.
ie when x = 12
so G is the point (12,-4).
P
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End of Unit 3
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