1. No category

NSM Enh 10_5.1-5.3 HB 02.fm Page 15 Wednesday, March 18, 2009 9:51 AM
Chapter
2
2:01
Quadratic Equations 1
Student Name
Class
Parent Signature
Date
Score
h (5 – x)(x – 2) = 0
Solution Using Factors
Outcome PAS5.3.2
A quadratic equation is one in which the highest
power of x is x2. One way of solving it is to use the
null factor law.
i
2
x(x + 3) = 0
Solve these quadratic equations.
a (x + 3)(2x – 3) = 0
When two quadratic factors multiply
together to give 0, either one factor
or the other factor is zero.
b (2x + 5)(x – 6) = 0
Quadratic solutions usually have two solutions.
Example 1 (already factorised):
Solve (x + 5)(x − 4) = 0.
Example 2 (complete the factorising step
yourself):
1
m
Solution:
x2 + 2x − 63 = 0
(x − 7)(x + 9) = 0
x − 7 = 0 or x + 9 = 0
x = 7, x = −9
Write down the solutions to these quadratic
equations.
Sa
a (x – 1)(x – 2) = 0
b (x + 4)(x – 3) = 0
c (x + 7)(x + 12) = 0
d (x – 1)(x + 13) = 0
s
pl
e
Solve x2 + 2x − 63 = 0.
d (4x + 1)(4x – 1) = 0
pa
ge
Solution:
Either x + 5 = 0 or x − 4 = 0
⇒ x = −5 or x = 4
The two solutions are −5 and 4.
c (2x – 1)(3x + 2) = 0
e (3x + 5)2 = 0
f
g 2x(x + 4) = 0
h 5x(1 – 8x) = 0
3
Solve these quadratic equations.
a x2 – 3x + 2 = 0
b x2 + 4x – 12 = 0
c x2 + 7x + 10 = 0
d x2 – x – 30 = 0
e x2 – 5x + 6 = 0
e (x + 4)2 = 0
f
f
(1 – x)(x + 7) = 0
✂
g x(x – 7) = 0
x(x – 3) = 0
x2 + x – 6 = 0
g x2 + 10x + 25 = 0
h x2 – 64 = 0
CHAPTER 2 QUADRATIC EQUATIONS
15
NSM Enh 10_5.1-5.3 HB 02.fm Page 16 Wednesday, March 18, 2009 9:51 AM
4
Solve these quadratic equations. Note that the
expressions contain common factors.
2:02
a x2 + 5x = 0
Solution by Completing
the Square
Outcome PAS5.3.2
Some quadratic equations can be written in the
form (x ± p)2 = q.
The process of changing an equation so that it
is written in this form is described as ‘completing
the square’.
The key step is to take half the co-efficient of x,
square it, and add the result to both sides of the
equation. The other steps are rearrangements.
b 3x2 – 3 = 0
c 3x2 – 7x = 0
Example: Solve x2 − 6x − 17 = 0.
Solution: x2 − 6x − 17 = 0
x2 − 6x = 17
x2 − 6x + 9 = 17 + 9 (key step)
(x − 3)2 = 26
x − 3 = ± 26
x = 3 ± 26
x = 3 ± 5.099
x = 8.099, x = −2.099
Quadratic equations can only be solved by
factorising if one side is 0.
The usual rules used when rearranging
equations apply.
Solution: x2 − x − 6 = 0
(x − 3)(x + 2) = 0
x = 3 or −2
Solve these quadratic equations by completing the
square. Round your answers correct to 2 dp if
they do not come out exactly.
a x2 – 12x + 3 = 0
Solve these quadratic equations.
b x2 + 18x – 1 = 0
Sa
a x2 + 4x = 45
m
5
1
pl
e
Example: Solve x2 = x + 6
pa
ge
s
d 5x2 – 10x – 40 = 0
b x2 = 2x + 15
c x2 + 9 = 6x
c x2 + x – 7 = 0
d x2 = 7x
d 3x2 + 4x – 5 = 0
16
NEW SIGNPOST MATHEMATICS ENHANCED 10 STAGE 5.1, 5.3 HOMEWORK BOOK
✂
e (x + 1)(x – 1) = 8
NSM Enh 10_5.1-5.3 HB 02.fm Page 17 Wednesday, March 18, 2009 9:51 AM
Chapter
2
2:03
Quadratic Equations 2
Student Name
Class
Parent Signature
Date
The Quadratic Formula
Score
2
Solve the equation x2 – 4x + 1 = 0, leaving your
answer in surd form.
3
Write the equation 10x + 7 = 4x2 in the form
ax2 + bx + c = 0, and then use the quadratic
formula to solve it correct to 2 dp.
Outcome PAS5.3.2
The two solutions (roots) of the quadratic
equation ax2 + bx + c = 0 are given by the formula:
–b ± b 2 – 4ac
x = -----------------------------------2a
Example: Solve the equation x2 + 7x − 3 = 0
Solution: In this equation a = 1, b = 7 and c = −3.
–7 ± 49 + 12
= ----------------------------------2
–7 ± 61 –7 ± 7.81
= ---------------------- = ---------------------2
2
x = 0.405 or −7.41
Use the quadratic formula to solve these quadratic
equations correct to 2 dp.
Sa
b x2 – 5x + 3 = 0
c 3x2 – 7x – 5 = 0
2:04
Choosing the Best Method
Outcome PAS5.3.2
m
a x2 + 18x – 4 = 0
pl
e
1
pa
ge
s
–7 ± 49 – 4 × 1 × – 3
x = ----------------------------------------------------2
When solving a quadratic equation, start by
expressing it in the form ax2 + bx + c = 0. You may
need to do some expanding and rearranging first.
Then factorise it if you can⎯otherwise use a
method like completing the square or use the
quadratic formula.
+1.
Example: Solve the equation x ( x + 2 ) = x----------3
Solution: First ‘clear’ the fraction:
x+1
x(x + 2) = ----------3
3x(x + 2) = x + 1
3x2 + 6x = x + 1
3x2 + 5x − 1 = 0
Use the quadratic formula with a = 3, b = 5
and c = −1:
–5 ± 25 – 4 × 3 × – 1
x = ----------------------------------------------------2
✂
–5 ± 37
= ---------------------2
x = 0.541 or −5.541
CHAPTER 2 QUADRATIC EQUATIONS
17
NSM Enh 10_5.1-5.3 HB 02.fm Page 18 Wednesday, March 18, 2009 9:51 AM
1
Solve these quadratic equations.
3
a x2 – 11x – 12 = 0
Solve these quadratic equations. You should
factorise first.
a 2x2 + 26x + 84 = 0
b 2x2 + 3x = 0
b x2 – x – 8 = 0
4
Solve these quadratic equations by expanding,
rearranging and then factorising.
pa
ge
s
a (x + 5)(x – 7) = 13
b (2x + 3)(x – 1) = x + 5
c 2x2 + 7x – 15 = 0
pl
e
c 4x + 5 = x(x – 2)
d (x + 1)(x – 7) = 3x – 7
Sa
m
d 3x2 + 5x – 8 = 0
2
5
Solve these quadratic equations.
12
x
a x + 1 = ------
Solve these quadratic equations by taking the
square root.
a x2 = 9
b (x – 3)2 = 36
14
x–3
b x + 2 = -----------
18
d
( x – 4 )2
------------------- = 18
2
NEW SIGNPOST MATHEMATICS ENHANCED 10 STAGE 5.1, 5.3 HOMEWORK BOOK
✂
c 16x2 = 25
NSM Enh 10_5.1-5.3 HB 02.fm Page 19 Wednesday, March 18, 2009 9:51 AM
Chapter
Quadratic Equations 3
2
2:05
Student Name
Class
Parent Signature
Date
Problems Involving
Quadratic Equations
1
Score
A right-angled triangle has a hypotenuse 15 cm
long and a base length x.
Outcome PAS5.3.2
When a practical situation results in a quadratic
equation, you should check each of the two
answers to see whether they are appropriate.
15
Example 1: One side of a rectangular garden is
3 m longer than the other side. The area of the
garden is 70 m2. Form a quadratic equation and
solve it to work out the dimensions of the garden.
x
a If the perimeter of the triangle is 36 cm, write
an expression for the height, h, of the triangle
in terms of x.
x
b If the area of the triangle is 54 cm2, form a
x⫹3
Solution:
pa
ge
x
s
x⫹3
x(x + 3) = 70
x2 + 3x = 70
2
x + 3x − 70 = 0
(x + 10)(x − 7) = 0
x = −10 or 7
pl
e
Because the question is about lengths we use
the positive answer only⎯that is, the shorter side
is 7 m, and the longer side is 7 + 3 = 10 m.
quadratic equation and solve it to find the
value of x.
2
A right-angled triangle has sides given by x,
3(x + 1) and 4x – 3. Use Pythagoras’ theorem to
calculate x if the hypotenuse is 4x – 3.
3
Michelle is six years younger than her sister, Lyn.
Four years ago the product of their ages was 40.
Form a quadratic equation and solve it to find the
ages of the sisters.
Sa
m
Example 2: A camper has two square tarpaulins.
Together they cover an area of 40 m2. The sides
of one sheet are each 4 m longer than the sides
of the other. Calculate the dimensions of the
smaller sheet.
x⫹4
x
Solution: x2 + (x + 4)2 = 40
x2 + x2 + 8x + 16 = 40
2x2 + 8x − 24 = 0
x2 + 4x − 12 = 0
(x + 6)(x − 2) = 0
x = −6, x = 2
✂
In this context the required length, x, is 2 m
and the smaller sheet measures 2 m by 2 m.
CHAPTER 2 QUADRATIC EQUATIONS
19
NSM Enh 10_5.1-5.3 HB 02.fm Page 20 Wednesday, March 18, 2009 9:51 AM
The cross-section of an
arch at the entrance to an
amusement park can be
modelled by the function
9 – x2 where x is the
distance in metres along
the ground from the centre
of the entrance.
4
9
6
–3
3
A dinghy has sunk and the survivors are on a life
raft. Their distance (measured in km) from a river
estuary is changing as the tide goes out and then
comes in again. It can be modelled by the equation
a How wide is the entrance at ground level?
d = –x2 + 3x + 4
b A float, which is 4 m high, will be driven
with x representing the number of hours since the
dinghy sank.
under the arch. The park’s engineers have
advised that the equation 9 – x2 = 4 is
involved. Solve this equation and explain what
the solution represents in this situation.
a Here is how some of the working for solving
the equation −x2 + 3x + 4 = 0 could be set out:
pa
ge
s
–x2 + 3x + 4 = 0
x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0
Write down the two solutions.
An open cistern is constructed from a square piece
of copper by removing squares measuring 7 cm by
7 cm from each corner. The sides are then folded
to form the cistern.
7 cm
7
7
A square photograph is to be cropped to fit into a
rectangular frame, which is 2 cm shorter in height
and 5 cm shorter in width. When this is done, the
area of the cropped photograph will be 54 cm2.
5
Sa
7
m
x
7
equation represents in this situation.
7
7
7
b Explain what the positive solution for this
pl
e
5
x
The cistern holds 9.1 litres⎯that is, its volume
is 9100 cm3. The problem is to calculate the
dimensions of the original piece of copper.
x
a Complete the 3-D diagram by adding 3 suitable
measurements in the correct positions.
b Complete the steps in this working:
2
length × width × height = volume
×
×
= 9100
= 9100
(x – 14)2 = 1300
c Solve the equation (x – 14)2 = 1300 and
x
Complete the solution of this equation to work
out the dimensions of the photograph before it
is cropped.
(x – 2)(x – 5) = 54
20
NEW SIGNPOST MATHEMATICS ENHANCED 10 STAGE 5.1, 5.3 HOMEWORK BOOK
✂
hence write down the dimensions of the
original piece of copper.