intermediate 2 – additional question bank unit 3 : further trig algebraic operations quadratic...
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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 3 :
FurtherTrig
AlgebraicOperations
Quadratic Functions
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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 3 : AlgebraicOperations
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1 2 3 4
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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)
as a single fraction in its simplest form.
ALGEBRAIC OPERATIONS : Question 1
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4
2 ( 7)
m m
m
Use the pattern
ab + c
dad + bc
bd=
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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
m2 - m
2m + 14=
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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)
as a single fraction in its simplest form.
7 4
10 ( 3)
t t
t
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ALGEBRAIC OPERATIONS: Question 1B
Express (t 3)
as a single fraction in its simplest form.
7 4
10 ( 3)
t t
t
Use the pattern
ab + c
dad + bc
bd=
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ALGEBRAIC OPERATIONS: Question 1B
Express (t 3)
as a single fraction in its simplest form.
7 4
10 ( 3)
t t
t
7t2 + 19t
10t – 30 =
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Question 1B
Express
(t 3)
as a single fraction in its
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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Comments
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
as a single fraction in its simplest form.
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ALGEBRAIC OPERATIONS: Question 2
2 2
3
2 15
5 4
a b
b a Express
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
2 2
3
2 15
5 4
a b
b a Express
as a single fraction in its simplest form.
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3b2a=
Question 2
Express
as a single fraction in its
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=
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2a2
5b x 15b2
4a3
30a2b2
20a3b=
3b2a=
1. Multiply top line then bottom line.
2. Cancel numbers then letters in alphabetical order.
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=
1. Multiply top line then bottom line.
2. Cancel numbers then letters in alphabetical order.
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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ALGEBRAIC OPERATIONS: Question 2B
2
3
2 6v v
w w Express
as a single fraction in its simplest form.
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ALGEBRAIC OPERATIONS: Question 2B
Express
as a single fraction in its simplest form.
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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ALGEBRAIC OPERATIONS: Question 2B
Express
as a single fraction in its simplest form.
2
3
2 6v v
w w vw2
3=
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Question 2B
Express
as a single fraction in its
simplest form.
1. To divide by a fraction : turn it upside down and multiply.
3. Cancel numbers then letters in alphabetical order.
1
3
w2v
2
3
2 6v v
w w
2v2
w 6v w3
2v2
w x 6vw3
=
2. Multiply top line then bottom line.
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.
3. Cancel numbers then letters in alphabetical order.
2v2
w 6v w3
2v2
w x 6vw3
=
2. Multiply top line then bottom line.
2v2w3
6vw=
vw2
3=
1
3
w2v
ab ÷ c
dadbc= a
b x dc =
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Comments
1. To divide by a fraction : turn it upside down and multiply.
3. Cancel numbers then letters in alphabetical order.
2v2
w 6v w3
2v2
w x 6vw3
=
2. Multiply top line then bottom line.
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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Deal with numbers and then apply laws
of indices: when dividing subtract
the powers.
Remember subtracting negative is like adding.
ALGEBRAIC OPERATIONS: Question 3
Simplify 5 3
4 424 8d d
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ALGEBRAIC OPERATIONS: Question 3
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.
ALGEBRAIC OPERATIONS: Question 3B
Simplify
5 2
3 3
2
a a
a
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ALGEBRAIC OPERATIONS: Question 3B
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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Learn Laws of Indices:
m n m na a a 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.
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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Comments
Learn Laws of Indices:
m n m na a a 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.
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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ALGEBRAIC OPERATIONS: Question 4
1 5 4
3 3 3m m m
Simplify giving your
answer with positive indices.
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ALGEBRAIC OPERATIONS: Question 4
1 5 4
3 3 3m m m
Simplify giving your
answer with positive indices.
Deal with numbers and then apply laws
of indices: when dividing subtract
the powers.
Remember subtracting negative is like adding.
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ALGEBRAIC OPERATIONS: Question 4
1 5 4
3 3 3m m m
Simplify giving your
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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Learn Laws of Indices:
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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ALGEBRAIC OPERATIONS: Question 4B
1 1 3
4 4 4w w w
Simplify
giving your answer without indices.
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Simplify
giving your answer without indices.
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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ALGEBRAIC OPERATIONS: Question 4B
Simplify
giving your answer without indices.
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
giving your answer without indices.
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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Comments
Learn Laws of Indices:
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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ALGEBRAIC OPERATIONS: Question 5
Evaluate 7c3/4 when c = 16
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Deal with indices first. Power ¾ is 4th root cubed.
Evaluate root before
power
ALGEBRAIC OPERATIONS: Question 5
Evaluate 7c3/4 when c = 16
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ALGEBRAIC OPERATIONS: Question 5
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
Learn Laws of Indices:
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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Comments
Learn Laws of Indices:
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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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
= 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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Comments
Learn Laws of Indices:
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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ALGEBRAIC OPERATIONS: Question 6
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.
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
X = 81
f(x) = 7x1/2 . Find the value of x if f(x) = 63.
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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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ALGEBRAIC OPERATIONS: Question 6B
f(x) = 2x1/3 . Find the value of x if f(x) = 20.
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Get hint
Equate both things that are equal to
f(x).
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.
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ALGEBRAIC OPERATIONS: Question 6B
X = 1000
f(x) = 2x1/3 . Find the value of x if f(x) = 20.
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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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Learn Laws of Indices:
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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ALGEBRAIC OPERATIONS: Question 7
Simplify 75 - 27 + 48
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Find the largest “perfect square” factor of each of
the numbers! Collect identical
surds in same way as you would
letters. .
ALGEBRAIC OPERATIONS: Question 7
Simplify 75 - 27 + 48
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ALGEBRAIC OPERATIONS: Question 7
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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Comments
Learn Laws of Indices:
1. Find the largest “perfect square” factor of each of the numbers!
2. Collect identical surds in same way as you would letters.
= 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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Comments
1. Find the largest “perfect square” factor of each of the numbers!
2. Collect identical surds in same way as you would letters.
= 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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ALGEBRAIC OPERATIONS: Question 7B
Simplify 80 + 45 - 180
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ALGEBRAIC OPERATIONS: Question 7B
Simplify 80 + 45 - 180
Find the largest “perfect square” factor of each of
the numbers! Collect identical
surds in same way as you would
letters. .
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ALGEBRAIC OPERATIONS: Question 7B
Simplify 80 + 45 - 180 = 5
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Question 7B
Simplify
80 + 45 - 180
1. Find the largest “perfect square” factor of each of the numbers!
2. Collect identical surds in same way as you would letters.
= 16 x 5 +
80 + 45 - 180
9 x 5 - 36 x 5
= 45 + 35 - 65
= 5 4x + 3x – 6x = x
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Learn Laws of Indices:
1. Find the largest “perfect square” factor of each of the numbers!
2. Collect identical surds in same way as you would letters.
= 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
1. Find the largest “perfect square” factor of each of the numbers!
2. Collect identical surds in same way as you would letters.
= 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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Deal with top and bottom lines separately. For each, find the
largest “perfect square” factor of
each of the numbers!
ALGEBRAIC OPERATIONS: Question 7C
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.
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ALGEBRAIC OPERATIONS: Question 7C
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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Learn Laws of Indices:
. .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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ALGEBRAIC OPERATIONS: Question 8
12
6Express 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.
ALGEBRAIC OPERATIONS: Question 8
12
6Express with a rational denominator.
a a a
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ALGEBRAIC OPERATIONS: Question 8
= 26
12
6Express with a rational denominator.
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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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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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Learn Laws of Surds:
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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ALGEBRAIC OPERATIONS: Question 8B
843
Simplify
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ALGEBRAIC OPERATIONS: Question 8B
Find the largest “perfect square” factor of each of
the numbers! Collect identical
surds in same way as you would
letters. .
843
Simplify
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ALGEBRAIC OPERATIONS: Question 8B
= 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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Learn Laws of Surds:
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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Learn Laws of Surds:
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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ALGEBRAIC OPERATIONS: Question 8C
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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To change the look of a fraction but not the value multiply top & bottom by the same
amount ie 2.
ALGEBRAIC OPERATIONS: Question 8C
When dealing with fractions check that you have
simplified as far as possible.
Find the value of tanx° giving your answer as a fraction with a rational denominator.
x°
3m
62m
0tanopp
xadj
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ALGEBRAIC OPERATIONS: Question 8C
Find the value of tanx° giving your answer as a fraction with a rational denominator.
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
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=
0tanopp
xadj
Simplify!!!1
4
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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
Learn Laws of Surds:
e.g.
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=
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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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 3 :Further Trig
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1 2
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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.
1200 2400 3600
y = asinbx°
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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.
1200 2400 3600
y = asinbx°
a = amplitude = ½ vertical extent. b = no. of waves in 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.
1200 2400 3600
y = asinbx°
a = 2 b = 3
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Question 1
1200 2400 3600
y = asinbx°
Write down the values of a and b.
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
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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
Learn basic trig graphs:
i.e. y = sinx˚
Max. = 1Min. = -1One cycle in 360˚
360
-1
1
x
y
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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
Learn basic trig graphs:
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° .
Write down the values of a and b.
y = acosbx°
900 1800 36002700
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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.
y = acosbx°
900 1800 36002700
a = amplitude = ½ vertical extent. b = no. of waves in 3600
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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.
y = acosbx°
900 1800 36002700
a = 5 b = 2
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y = acosbx°
900 1800 36002700
Question 1B
Write down the values of a and b.
1. a = amplitude = ½ vertical extent.
max /min = ±5 so a = 5
2. b = no. of waves in 3600
2 complete waves from 0 to 360
so b = 2
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1. a = amplitude = ½ vertical extent.
max /min = ±5 so a = 5
2. b = no. of waves in 3600
2 complete waves from 0 to 360
so b = 2
Learn basic trig graphs:
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
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
Learn basic trig graphs:
i.e. y = cosx˚
y = acosx˚
Stretch factor
y = cosbx˚
Number of cycles in 360˚
One cycle
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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
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.
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Further Trig : Question 2
Solve 5sinx° - 1 = 1 where 0<x<360
x = 23.6°
x = 156.4°
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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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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 = …...
3. Calculate angles remembering that answers to trig equations come in pairs.
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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Further Trig : Question 2B
Solve 3tanx° + 7 = 2 where 0<x<360
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Further Trig : Question 2B
Solve 3tanx° + 7 = 2 where 0<x<360
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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Solve 3tanx° + 7 = 2 where 0<x<360
Further Trig : Question 2B
x = 121.0°
x = 301.0°
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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
1. Always re-arrange to tanx0 = …...
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°
3. Calculate angles remembering that answers to trig equations come in pairs.
= 301.0°
= 121.0°
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1. Always re-arrange to tanx0 = …...
tanx = -5/3
3tanx° + 7 = 2
3tanx° = -5
tan-1 (53) = 59.0
Q2: angle = 180° - 59.0°
Q4: angle = 360° - 59.0°
3. Calculate angles remembering that answers to trig equations come in pairs.
= 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
1. Always re-arrange to tanx0 = …...
tanx = -5/3
3tanx° + 7 = 2
3tanx° = -5
tan-1 (53) = 59.0
Q2: angle = 180° - 59.0°
Q4: angle = 360° - 59.0°
3. Calculate angles remembering that answers to trig equations come in pairs.
= 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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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 3 : Quadratic Functions
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Please choose a question to attempt from the following:
1 2
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3 4
Quadratic Functions: Question 1
Solve the equation x2 – 6x + 7 = 0 giving your answers to 2 decimal places.
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Quadratic Functions: Question 1
Solve the equation x2 – 6x + 7 = 0 giving your answers to 2 decimal places.
Write down values of a,
b & c. Evaluate b2 – 4ac .
Now use quadratic formula,
remembering that you should get
two answers.
Remember to round.
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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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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
to 2 decimal places.
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
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
to 2 decimal places.
1. Write down values of a, b & c.
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
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
24. 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!!
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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
24. Rewrite with brackets and now use
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
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
24. Rewrite with brackets and now use
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
1. Write down values of a, b & c.
x2 – 6x + 7 = 0
a = 1 b = -6 c = 7
b2 – 4ac
2. Evaluate b2 – 4ac .
b2 – 4ac = (-6)2 – (4x1x7)
= 36 – 28
= 8
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Quadratic Functions: Question 1B
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Solve the equation 3x2 + 5x - 1 = 0 giving your answers to 2 significant figures.
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Quadratic Functions: Question 1B
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Write down values of a,
b & c. Evaluate b2 – 4ac .
Now use quadratic formula,
remembering that you should get
two answers.
Remember to round.
Solve the equation 3x2 + 5x - 1 = 0 giving your answers to 2 significant figures.
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Quadratic Functions: Question 1B
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x = 0.18 or -1.8
Solve the equation 3x2 + 5x - 1 = 0 giving your answers to 2 significant figures.
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Question 1B
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1. Write down values of a, b & c.
3x2 + 5x – 1 = 0
a = 3 b = 5 c = (-1)
2. Evaluate b2 – 4ac .
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.
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Question 1B
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1. Write down values of a, b & c.
3. Now use quadratic formula.
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)
Solve the equation
3x2 + 5x - 1 = 0
giving your answers
to 2 significant figures.
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Question 1B
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1. Write down values of a, b & c.
4. Rewrite with brackets and now use calculator.
= (-5 + 37) 2 or (-5 - 37) 2
5. Remember to round.
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
to 2 significant figures.
= 0.180… or -1.847..
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Comments
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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!!
1. Write down values of a, b & c.
3. Now use quadratic formula.
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
Refer to the Formula Sheet:
x = -b ± (b2 – 4ac )
2a
Quadratic Formula:
1. Write down values of a, b & c.
3. Now use quadratic formula.
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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Take care when allocating a
value to a, b and c.
ax2 + bx + c = 0
3x2 + 5x - 1 = 0
a =3b = 5c = -1
1. Write down values of a, b & c.
3. Now use quadratic formula.
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)
Note:
Finding first eases working.
Bracket all negative numbers. Watch out for double negative!
Comments
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b2 – 4ac1. Write down values of a, b & c.
3x2 + 5x – 1 = 0
a = 3 b = 5 c = (-1)
2. Evaluate b2 – 4ac .
b2 – 4ac = (5)2 – (4 x 3 x -1)
= 25 – (- 12)
= 37
Quadratic Functions: Question 2
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Reveal answer only
EXIT
Get hint
Solve the equation 3x2 = x + 1 giving your answers to 2 decimal places.
What would you like to do now?
Quadratic Functions: Question 2
Go to full solution
Go to Comments
Go to Quadratics Menu
Reveal answer only
EXIT
Write down values of a,
b & c. Evaluate b2 – 4ac .
Now use quadratic formula,
remembering that you should get
two answers.
Remember to round.
Solve the equation 3x2 = x + 1 giving your answers to 2 decimal places.
Re-write quadratic:
make it equal to
zero before solving.
What would you like to do now?
Quadratic Functions: Question 2
Go to full solution
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EXIT
Try another like this
x = 0.77 or -0.43
Solve the equation 3x2 = x + 1 giving your answers to 2 decimal places.
What would you like to do now?
Comments
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Question 2
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2. Write down values of a, b & c.
3x2 – x – 1 = 0
a = 3 b = (-1) c = (-1)
3. Evaluate b2 – 4ac .
b2 – 4ac = (-1)2 – (4 x 3 x -1)
= 1 – (-12)
= 13Continue Solution
Solve the equation
3x2 = x + 1
giving your answers
to 2 decimal places.
1. Rearrange in quadratic form.
3x2 = x + 1
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Question 2
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2. Write down values of a, b & c.
3x2 –x – 1 = 0
a = 3 b = (-1) c = (-1)
3. Now use quadratic formula.
x = -b ± (b2 – 4ac )
2a
x = 1 ± 13
24. Rewrite with brackets and now use
calculator.
= (1 + 13) 2 or (1 - 13) 2
Solve the equation
3x2 = x + 1
giving your answers
to 2 decimal places.
Comments
Begin Solution
Question 2
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1. Write down values of a, b & c.
3x2 – x – 1 = 0
a = 3 b = (-1) c = (-1)
4. Rewrite with brackets and now use calculator.
= (1 + 13) 2 or (1 - 13) 2
5. Remember to round.
x = 0.77 or -0.43Try another like this
Solve the equation
3x2 = x + 1
giving your answers
to 2 decimal places.
= 0.767… or -0.434..
What would you like to do now?
Comments
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Next Comment
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!!
2. Write down values of a, b & c.
3x2 – x – 1 = 0
a = 3 b = (-1) c = (-1)
3. Evaluate b2 – 4ac .
b2 – 4ac = (-1)2 – (4 x 3 x -1)
= 1 – (-12)
= 13
1. Rearrange in quadratic form.
3x2 = x + 1
Comments
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Next Comment
You must put the quadratic into standard quadratic form (make equation equal zero) before attempting to solve.
2. Write down values of a, b & c.
3x2 – x – 1 = 0
a = 3 b = (-1) c = (-1)
3. Evaluate b2 – 4ac .
b2 – 4ac = (-1)2 – (4 x 3 x -1)
= 1 – (-12)
= 13
1. Rearrange in quadratic form.
3x2 = x + 1
Comments
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ax2 + bx + c = 0
Refer to the Formula Sheet:
x = -b ± (b2 – 4ac )
2a
Quadratic Formula:
2. Write down values of a, b & c.
3x2 –x – 1 = 0
a = 3 b = (-1) c = (-1)
3. Now use quadratic formula.
x = -b ± (b2 – 4ac )
2a
x = 1 ± 13
24. Rewrite with brackets and now
use calculator.
= (1 + 13) 2 or (1 - 13) 2
Comments
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Take care when allocating a
value to a, b and c.
ax2 + bx + c = 0
3x2 - 1x - 1 = 0
a =3b = (-1)c = (-1)
2. Write down values of a, b & c.
3x2 –x – 1 = 0
a = 3 b = (-1) c = (-1)
3. Now use quadratic formula.
x = -b ± (b2 – 4ac )
2a
x = 1 ± 13
24. Rewrite with brackets and now
use calculator.
= (1 + 13) 2 or (1 - 13) 2
Note:
Finding first eases working.
Bracket all negative numbers. Watch out for double negative!
Comments
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b2 – 4ac
2. Write down values of a, b & c.
3x2 – x – 1 = 0
a = 3 b = (-1) c = (-1)
3. Evaluate b2 – 4ac .
b2 – 4ac = (-1)2 – (4 x 3 x -1)
= 1 – (-12)
= 13
1. Rearrange in quadratic form.
3x2 = x + 1
Try another like this
Quadratic Functions: Question 2B
Go to full solution
Go to Comments
Go to Quadratics Menu
Reveal answer only
EXIT
Get hint
Solve the equation 2x(x + 2) = 2 - x
giving your answers to 2 significant figures.
What would you like to do now?
Quadratic Functions: Question 2B
Go to full solution
Go to Comments
Go to Quadratics Menu
Reveal answer only
EXIT
Write down values of a,
b & c.
Evaluate b2 – 4ac .
Now use quadratic formula,
remembering that you should get
two answers.
Remember to round.
Solve the equation 2x(x + 2) = 2 - x
giving your answers to 2 significant figures.
Re-write quadratic: get rid of
brackets & make it equal to
zero before solving.
What would you like to do now?
Quadratic Functions: Question 2B
Go to full solution
Go to Comments
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EXIT
x = 0.35 or -2.9
Solve the equation 2x(x + 2) = 2 - x
giving your answers to 2 significant figures.
What would you like to do now?
Comments
Begin Solution
Continue Solution
Question 2B
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Solve the equation
2x(x + 2) = 2 - x
giving your answers
to 2 significant figures.
2. Write down values of a, b & c.
2x2 + 5x – 2 = 0
a = 2 b = 5 c = (-2)
3. Evaluate b2 – 4ac .
b2 – 4ac = (5)2 – (4 x 2 x -2)
= 25 – (-16)
= 41
1. Rearrange in quadratic form.
2x(x + 2) = 2 – x 2x2 + 4x – 2 + x= 0
Comments
Begin Solution
Continue Solution
Question 2B
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3. Now use quadratic formula.
x = -b ± (b2 – 4ac )
2a
x = -5 ± 41
24. Rewrite with brackets and now use
calculator.
= (-5 + 41) 2 or (-5 - 41) 2
Solve the equation
2x(x + 2) = 2 - x
giving your answers
to 2 significant figures.
2x2 + 5x – 2 = 0
a = 2 b = 5 c = (-2)
1. Rearrange in quadratic form.
Comments
Begin Solution
Continue Solution
Question 2B
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4. Rewrite with brackets and now use calculator.
= (-5 + 41) 2 or (-5 - 41) 2
5. Remember to round.
x = 0.35 or -2.9
2x2 + 5x – 2 = 0
a = 2 b = 5 c = (-2)
1. Rearrange in quadratic form.
Solve the equation
2x(x + 2) = 2 - x
giving your answers
to 2 significant figures.
= 0.350… or -2.850..
What would you like to do now?
Comments
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Next Comment
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!!
2. Write down values of a, b & c.
2x2 + 5x – 2 = 0
a = 2 b = 5 c = (-2)
3. Evaluate b2 – 4ac .
b2 – 4ac = (5)2 – (4 x 2 x -2)
= 25 – (-16)
= 41
1. Rearrange in quadratic form.
2x(x + 2) = 2 – x 2x2 + 4x – 2 + x= 0
Comments
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2. Write down values of a, b & c.
2x2 + 5x – 2 = 0
a = 2 b = 5 c = (-2)
3. Evaluate b2 – 4ac .
b2 – 4ac = (5)2 – (4 x 2 x -2)
= 25 – (-16)
= 41
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.
Comments
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ax2 + bx + c = 0
Refer to the Formula Sheet:
x = -b ± (b2 – 4ac )
2a
Quadratic Formula:
3. Now use quadratic formula.
x = -b ± (b2 – 4ac )
2a
x = -5 ± 41
24. Rewrite with brackets and now
use calculator.
= (-5 + 41) 2 or (-5 - 41) 2
2x2 + 5x – 2 = 0
a = 2 b = 5 c = (-2)
1. Rearrange in quadratic form.
Comments
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Next Comment
Take care when allocating a
value to a, b and c.
ax2 + bx + c = 0
2x2 + 5x - 2 = 0
a =2b = 5c = -2
3. Now use quadratic formula.
x = -b ± (b2 – 4ac )
2a
x = -5 ± 41
24. Rewrite with brackets and now
use calculator.
= (-5 + 41) 2 or (-5 - 41) 2
2x2 + 5x – 2 = 0
a = 2 b = 5 c = (-2)
1. Rearrange in quadratic form.
Note:
Finding first eases working.
Bracket all negative numbers. Watch out for double negative!
Comments
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b2 – 4ac
2. Write down values of a, b & c.
2x2 + 5x – 2 = 0
a = 2 b = 5 c = (-2)
3. Evaluate b2 – 4ac .
b2 – 4ac = (5)2 – (4 x 2 x -2)
= 25 – (-16)
= 41
1. Rearrange in quadratic form.
2x(x + 2) = 2 – x 2x2 + 4x – 2 + x= 0
Quadratic Functions: Question 3
Go to full solution
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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.
What would you like to do now?
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.
Quadratic Functions: Question 3
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
Go to Quadratics Menu
Reveal answer
What would you like to do now?
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.
Quadratic Functions: Question 3
EXIT
Try another like this
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
(a) Show that the total area is given by the expression 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.
Try another like this
What would you like to do now?
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2. Use the expression for area given in part (a) and make it equal to 60.
If total area = 60 (b)
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
Length must be 5m as negative
value not valid.
3. Make the equation equal to zero, factorise and solve.
The quadratic equation can also be solved by applyingthe quadratic formula:
ax2 + bx + c = 0
Refer to the Formula Sheet:
x = -b ± (b2 – 4ac )
2a
Quadratic Formula:
Finding first eases working.Bracket all negative numbers.
b2 – 4ac
Comments
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2. Use the expression for area given in part (a) and make it equal to 60.
If total area = 60 (b)
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
Length must be 5m as negative
value not valid.
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 + 7x - 60 = 0
a =1b = 7c = (-60)
Comments
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2. Use the expression for area given in part (a) and make it equal to 60.
If total area = 60 (b)
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
Length must be 5m as negative
value not valid.
3. Make the equation equal to zero, factorise and solve.
Make sure that you understand when a
negative answer may not be
correct in context given,
e.g. negative lengths or areas.
Try another like this
Quadratic Functions: Question 3B
Go to full solution
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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.
(b) Hence find the value of x
to meet this requirement.
(a)Find an expression for the area of the garage in terms of x.
What would you like to do now?
(b) Hence find the value of x
to meet this requirement.
(a)Find an expression for the area of the garage in terms of x.
Quadratic Functions: Question 3B
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
Go to Quadratics Menu
Reveal answer
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.
What would you like to do now?
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.
(b) Hence find the value of x
to meet this requirement.
(a)Find an expression for the area of the garage in terms of x.
Quadratic Functions: Question 3B
EXIT Go to full solution
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= 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.
Comments
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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
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 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
3. Make the equation equal to zero, factorise and solve.
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Question 3B
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(b)
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.
x2 + 4x - 21 = 0
(x + 7)(x – 3) = 0
(x + 7) = 0 or (x – 3) = 0
3. Make the equation equal to zero, factorise and solve.
ie. x = -7 or x = 3
Length must be 3m as
negative value not valid.
What would you like to do now?
Comments
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Next Comment
The quadratic equation can also be solved by applyingthe quadratic formula:
ax2 + bx + c = 0
Refer to the Formula Sheet:
x = -b ± (b2 – 4ac )
2a
Quadratic Formula:
Finding first eases working.Bracket all negative numbers.
b2 – 4ac
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
= 21m2
So x2 + 4x = 21
x2 + 4x - 21 = 0
(x + 7)(x – 3) = 0
(x + 7) = 0 or (x – 3) = 0
3. Make the equation equal to zero, factorise and solve.
Comments
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Take care when allocating a
value to a, b and c.
ax2 + bx + c = 0
1x2 + 4x - 21 = 0
a =1b = 4c = (-21)
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
= 21m2
So x2 + 4x = 21
x2 + 4x - 21 = 0
(x + 7)(x – 3) = 0
(x + 7) = 0 or (x – 3) = 0
3. Make the equation equal to zero, factorise and solve.
Comments
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Next Comment
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
3. Make the equation equal to zero, factorise and solve.
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.
Quadratic Functions: Question 4
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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
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.
In ‘completed square’, a
minimum value occurs when
bracket is zero.
Quadratic Functions: Question 4
EXIT
Use symmetry of a parabola to find other root if you know one.
Go to full solution
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Go to Quadratics Menu
Reveal answer
The blades in electric hair-clippers have grooves in the form of identical adjacent parabolas.
X
Y
E F G
C D
roots are equidistant from axis of symmetry.
use ‘completed square’ form to write equation of parabola.
What would you like to do now?
(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.
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
Quadratic Functions: Question 4
EXIT
Try another like this
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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
E is the point (5,-9).
(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
E is the point (5,-9).
(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.
What would you like to do now?
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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).
1. In ‘completed square’, a minimum value occurs when bracket is zero.
(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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Learn the following Results:
y = (x – 5)2 - 9
Minimum T.P. at P(5,-9)
e.g.
P
1. In ‘completed square’, a minimum value occurs when bracket is zero.
(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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Try another like this
Learn the following Results:
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).
Quadratic Functions: Question 4B
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Go to CommentsReveal answer
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.
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
Quadratic Functions: Question 4B
EXIT Go to full solution
Go to Comments
Go to Quadratics Menu
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.
use ‘completed square’ form to write equation of parabola.
What would you like to do now?
(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.
Quadratic Functions: Question 4B
EXIT
Try another like this
Go to full solution
Go to Comments
Go to Quadratics Menu
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.
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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Continue Solution
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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Continue Solution
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
U is the point (3,4).
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)
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)2Try another like this
(c) Find the equation of the
parabola with maximum W.
U V W
X
YT
U is the point (3,4).
What would you like to do now?
Comments
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Back to Home
Next Comment
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).
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).
Comments
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Learn the following Results:
y = 4 – (x – 3)2
Maximum T.P. at P(3,4)
e.g.
P
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).
Try another like this
(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
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Go to CommentsReveal answer
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
(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.
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 . In ‘completed
square’, a minimum value
occurs when bracket is zero.
Quadratic Functions: Question 4C
EXIT
If not given a root you must solve the quadratic.
Go to full solution
Go to Comments
Go to Quadratics Menu
Reveal answer
use ‘completed square’ form to write equation of parabola.
The diagram below shows some identical adjacent parabolas.
X
Y
F G
H J
What would you like to do now?
(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
EXIT Go to full solution
Go to Comments
Go to Quadratics Menu
G is the point (12,-4).
y = (x – 8)2 - 4
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
H is (10,0) and J is (14,0)
What would you like to do now?
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Question 4C
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Continue Solution
1. In ‘completed square’, a minimum value occurs when bracket is zero.
(a)
y will have a minimum value of
-4 when (x – 12)2 = 0.
ie when x = 12
so G is the point (12,-4).
X
Y
F G
H J
Its equation is y = (x – 12)2 – 4 .
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Question 4C
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Continue Solution
2. If you are not given one of the roots you must solve quadratic to find roots.
G is the point (12,-4).
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.
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
(c) Find the equation of the
parabola with minimum F.
G is the point (12,-4).
10 14
So F is the point (8,-4).
ie y = (x – 8)2 - 4
Continue Solution
What would you like to do now?
Comments
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Next Comment
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).
1. In ‘completed square’, a minimum value occurs when bracket is zero.
(a)
y will have a minimum value of
-4 when (x – 12)2 = 0.
ie when x = 12
so G is the point (12,-4).
Comments
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Next Comment
Learn the following Results:
y = (x – 12)2 – 4
Minimum T.P. at P(12,-4)
e.g.
P
1. In ‘completed square’, a minimum value occurs when bracket is zero.
(a)
y will have a minimum value of
-4 when (x – 12)2 = 0.
ie when x = 12
so G is the point (12,-4).
End of Unit 3