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Quadratic Equations
Curriculum Ready
Quadratic Equations
ACMNA: 233, 269
Solutions
1100% Quadratic Equations Solutions
Mathletics 100% © 3P Learning
Quadratic Equations
SERIES TOPIC
K 7
Solutions Basics
Page 3 questions
1. What is the difference between a linear and a quadratic equation?
2. Find the solutions to the variable in each of these equations. Leave the solutions in surd form if necessary, and state if each equation has one, two or no real solutions:
a b c
d e f
g h i
j
x
x
x
4
2
2
2
2 2!
!
=
=
=
^ h
2 32b
b
b
b
16
4
4
2
2
2 2!
!
=
=
=
=
^ h
3 75 0b
b
b
b
25
5
5
2
2
2 2!
!
- =
=
=
=
^ h
6 0y
y
y
0
0
2
2
=
=
=
t
t
t
t
4 28 0
428
7
7
2
2
2 2!
!
- =
=
=
=
^ h
9h2=-
2 8
4
p
p
2
2
=-
=-
8 5 5m
m
m
8 0
0
2
2
+ =
=
=
x
x
x
x
9 16 0
916
34
34
2
2
2 2!
!
- =
=
=
=
` j
3 108 0k
k
k
k
36
6
6
2
2
2 2!
!
- + =
=
=
=
^ h
In a linear equation the highest index of the variable (eg x ) has an index of 1. In a quadratic equation the highest index of the variable has an index of 2 (eg x2 ).
No real solutions
One solution
One solution
No real solutions
Two solutions
Two solutions
Two solutions
Two solutions
Two solutions
Two solutions
2 100% Quadratic Equations Solutions
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Quadratic Equations
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1. Rewrite these polynomials in the form 0ax bx c2+ + = . Identify the values of a, b and c:
a
a
c
c
e
e
g
b
b
d
d
f
f
h
2. Solve these quadratic equations for the missing variable:
3 2 0x x2+ + =
1, 3, 2a b c` = = =
7 0x
x x0 7 0
2
2
- =
+ - =
1, 0, 7a b c` = = =-
3 0x x
x x
4 5
12 15 0 02
- =
- + =
^ h
12, 15, 0a b c` = =- =
1, 4, 21a b c` = =- =-
0x x
x x x
x x
3 5 8
3 5 24 40 0
3 19 40 0
2
2
+ - =
+ - - =
- - =
^ ^h h
3, 19, 40a b c` = =- =-
2 4 5 0x x2+ + =
2, 4, 5a b c` = = =
x x
x x
4 0
4 0 02
+ =
+ + =
^ h
1, 4, 0a b c` = = =
0x x
x x x
x x
3 7
3 7 21 0
4 21 0
2
2
+ - =
+ - - =
- - =
^ ^h h
x x
x x x
x x
3 4 1 0
3 4 4 0
3 9 12 0
2
2
- + - =
- + - - =
- - + =
^ ^
^
h h
h
3, 9, 12a b c` =- =- =
0x x 2- =^ h
0x x2 1+ =^ h
0y y2 3- =^ h
0x x 4+ =^ h
0x x3 5- + =^ ^h h
0x x2 7 3 8+ - =^ ^h h
Page 5 questions
x 0=
x 0=
0y = x2 7 0+ =
x 3 0- =
x 0=
x 0=
x 0=
0y = x27=-
x 3=
x 0=
x 2 0- =
2 1 0x + =
y2 3 0- = x3 8 0- =
x 5 0+ =
0x 4+ =or
or
or or
or
or
or
or
or or
or
or
(By Null Factor Law) (By Null Factor Law)
(By Null Factor Law) (By Null Factor Law)
(By Null Factor Law) (By Null Factor Law)
x 2=
x21=-
y23= x
38=
5x =-
x 4=-
3100% Quadratic Equations Solutions
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Page 6 questions
Page 7 questions
x 0=
0t =
y 7 0- =
n 8 0+ = x 1 0- =
m 6 0- =
0b5 =
0x4 =
x 0=
0t =
y 7=
n 8=- x 1=
m 6=
0b =
x 0=
0x 8+ =
0t 6- =
y 3 0+ =
n 4 0- = x 6 0+ =
m 5 0+ =
0b 3- =
0x 3+ =or
or
or
or or
or
or
or
or
or
or
or or
or
or
or
(By Null Factor Law) (By Null Factor Law)
(By Null Factor Law) (By Null Factor Law)
(By Null Factor Law)
(By Null Factor Law)
(By Null Factor Law)
(By Null Factor Law)
x 8=-
6t =
y 3=-
n 4= x 6=-
m 5=-
b 3=
x 3=-
3. Solve these quadratic equations by factorising:
a
c
e
b
d
f
8 0
0
x x
x x 8
2+ =
+ =^ h
6 0
0
t t
t t 6
2- =
- =^ h
4 21 0
0
y y
y y7 3
2- - =
- + =^ ^h h
4 12 0
4 0
x x
x x 3
2+ =
+ =^ h
5 15 0
5 0
b b
b b 3
2- =
- =^ h
30 0
0
m m
m m6 5
2- - =
- + =^ ^h h
n n
n n
n n
2 8 64 0
2 4 32 0
2 8 4 0
2
2
+ - =
+ - =
+ - =
^
^ ^
h
h h
g h 5 6x x
x x
x x
5 6 0
1 6 0
2
2
+ =
+ - =
- + =^ ^h h
4 100% Quadratic Equations Solutions
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i j3 105x x
x x
x x
x x
2
3 6 105 0
3 2 35 0
3 7 5 0
2
2
+ =
+ - =
+ - =
+ - =
^
^
^ ^
h
h
h h
8 72p p
p p p
p p
p p
3
6 9 8 72
2 63 0
9 7 0
2
2
2
+ = +
+ + = +
- - =
- + =
^
^ ^
h
h h
Page 7 questions
Page 9 questions
x 7 0+ = p 9 0- =
x 7=- p 9=
x 5 0- = p 7 0+ =or or
or or
(By Null Factor Law) (By Null Factor Law)
x 5= p 7=-
4. Complete the square of the following trinomials:
a bx x x
x
4 1 2 2 1
2 3
2 2 2
2
+ + = + - +
= + -
^
^
h
h
x x x
x
6 16 3 3 16
3 7
2 2 2
2
- + = - - - +
= - +
^ ^
^
h h
h
c
e
d
f
x x x
x
x
x
3 723
23 7
23
49 7
23
49
428
23
419
22 2
2
2
2
+ + = + - +
= + - +
= + - +
= + +
` `
`
`
`
j j
j
j
j
x x x x
x
x
x
3 24 30 3 8 10
3 4 4 10
3 4 6
3 4 18
2 2
2 2
2
2
+ + = + +
= + - +
= + -
= + -
^
^
^
h
h
h
6
6
6
@
@
@
x x x x
x
x
x
10 23 10 23
5 5 23
5 2
5 2
2 2
2 2
2
2
- + - =- - +
=- - - - +
=- - -
=- - +
^ ^
^
^
h h
h
h
6
6
6
@
@
@
x x x x
x
x
x
3 6 18 3 2 6
3 1 1 6
3 1 7
3 1 21
2 2
2 2
2
2
- - + =- + -
=- + - -
=- + -
=- + +
^
^
^
h
h
h
6
6
6
@
@
@
5100% Quadratic Equations Solutions
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Page 11 questions
x 2 5+ = x 3 4= -
x 2 5=- +
x 7 232= +
x 3 7=- + x 5 10= +
7 2x32= -
x 3 7=- - x 5 10= -
1x =-or or
x 2 5+ =- x 3 4= +or
or
or or
or
x 2 5=- - x 7=
5. Solve for x in the following:
a
a
c
b
b
d
x 2 52+ =^ h
x
x
2 5
2 5
2 2!
!
+ =
+ =
^ ^h h
x
x
x
x
3 3 2 0
3 7
3 7
3 7
2 2
2
2 2!
!
+ - + =
+ =
+ =
+ =
^
^
^ ^
h
h
h h
x
x
x
x
5 5 15 0
5 10
5 10
5 10
2 2
2
2 2!
!
- - - + =
- =
- =
- =
^ ^
^
^ ^
h h
h
h h
x
x
3 4
3 4
22 !
!
- =
- =
^ ^h h
16x 3 2- =^ h
x
x
x
x
3 7 8
738
738
73
4 2
2
2
2
2
!
! #
- =
- =
- =
- =
^
^
c
c
h
h
m
m
5 35
7
x
x
6
6
2
2
- + =
+ =-
^
^
h
h
6. Complete the square and then solve for x in the following:
6 2 0x x2+ + = 10 15 0x x2
- + =
No real solution
6 100% Quadratic Equations Solutions
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Quadratic Equations
SERIES TOPIC
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Knowing MoreSolutions
7. Complete the square and solve for the variable in the following:
a
c
b
d
2 3 2 0q q2+ - = 2 5 0m m2
- - + =
4 8 1 0t t2+ - = 3 12 2 0x x2
- + - =
Page 12 questions
q q
q
q
q
q
223 1 0
243
43 1 0
243
169
1616 0
43
1625 0
43
45
2
2 2
2
2
2 2!
+ - =
+ - - =
+ - - =
+ - =
+ =
` `
` ` `
`
` `
j j
j j j
j
j j
8
;
;
B
E
E
q43
45=- +
1m 6=- +
1t45=- + x 2
310= +
q21=
q43
45=- -
m 1 6=- -
1t45=- -
310x 2= -
q 2=-
or
or
or or
or
m m
m
m
m
m
2 5 0
1 1 5 0
1 6 0
1 6
1 6
2
2 2
2
2 2!
!
- + - =
- + - - =
+ - =
+ =
+ =
^
^
^ ^
h
h
h h
6
6
@
@
t t
t
t
t
t
4 241 0
4 1 141 0
145 0
145
145
2
2 2
2
22
!
!
+ - =
+ - - =
+ - =
+ =
+ =
^
^
^ c
h
h
h m
8
8
B
B
x x
x
x
x
x
3 432 0
3 2 232 0
2 432 0
2312
32
2310
2
2 2
2
2
22
!
- - + =
- - - - + =
- - + =
- = -
- =
^ ^
^
^
^ c
h h
h
h
h m
8
8
B
B
Divide both sides by -3
Divide both sides by -1Divide both
sides by 2
Divide both sides by 4
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Page 13 questions
8. Consider the equation 0ax bx c2+ + = where a, b and c are any constants:
a
b
Complete the square for this equation
Solve for x
ax bx c
a xab x
ac
a xab
ab
ac
xab
ab
ac
xab
a
bac
0
0
2 20
2 20
2 4
2
2
2 2
2 2
2
2
2
+ + =
+ + =
+ - + =
+ - + =
+ = -
` `
` `
`
j j
j j
j
8
;
B
E
xab
a
bac
xab
a
b ac
xab
ab ac
xa
b b ac
2 4
2 4
4
2 24
24
2
2
2
2
2
2
2
!
!
!
+ = -
+ = -
= - -
= - -
` j
Divide both sides by a
8 100% Quadratic Equations Solutions
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Page 15 questions
1. Use the quadratic formula to solve the following quadratic equations, leaving your answers in surd form:
a
c
b
d
3 18 0x x2- - = 9 14x x 02
- + =
2 26 80 0x x2+ + = 4 32 60 0t t2
- + - =
xa
b b ac
x
x
x
x
24
2 1
3 3 4 1 18
23 9 72
23 81
23 9
2
2
!
#
! # #
!
!
!
= - -
=- - - - -
= +
=
=
^ ^h h
1, 3, 18a b c= =- =-
x 6=x 7=
3x =-x 2=
or
or
or
or
xa
b b ac
x
x
x
x
24
2 1
9 9 4 1 14
29 81 56
29 25
29 5
2
2
!
#
! # #
!
!
!
= - -
=- - - -
= -
=
=
^ ^h h
1, 9, 14a b c= =- =
xa
b b ac
x
x
x
24
2 226 26 4 2 80
426 676 640
426 36
2
2
!
#! # #
!
!
= - -
= - -
= - -
= -
8x =-
x 5=
5x =-
x 3=
2, 26, 80a b c= = =
4
32 4 60
832
32
832 8
xa
b b ac
x
x
x
x
24
2
32 4
1024 960
864
2
2
!
!
!
!
!
= - -
=-
- - - -
=-
- -
=-
-
=-
-
^^ ^hh h
4, 32, 60a b c=- = =-
9100% Quadratic Equations Solutions
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Page 15 questions
Page 16 questions
e f2 5 0y y2+ - = 5 0p p 52
- - + =
or
or
or
orx 1 6=- +
x 4=-
x23=
x 1 6=- -
x32=
x23=-
2 5
2
1
xa
b b ac
x
x
x
x
24
2 1
2 4 1
24 20
424
1 6
2
2
!
!
!
!
!
= - -
=- - -
= - +
=-
=-
^^ ^hh h
1, 2, 5a b c= = =-
xa
b b ac
x
x
x
x
24
2 1
5 5 4 1 5
25 25 20
25 45
25
25 9
2
2
!
!
!
!
! #
= - -
=-
- - - - -
=-
+
=-
=--
^^ ^ ^ ^
hh h h h
x25
23 5=- + x
25
23 5=- -
1, 5, 5a b c=- =- =
2. Use the quadratic formula to solve the following quadratic equations, leaving your answers in surd form:
a b3 10 8 0x x2- - + = 4 9 0x2
- =
3
10 10 3
6
xa
b b ac
x
x
x
24
2
4 8
10 100 96
610 196
2
2
!
!
!
!
= - -
=-
- - - - -
=-
+
=-
^^ ^ ^ ^
hh h h h
3, 10, 8a b c=- =- =
xa
b b ac
x
x
24
2 4
0 0 4 4 9
8144
2!
!
!
= - -
=- -
=
^^ ^hh h
4, 0, 9a b c= = =-
10 100% Quadratic Equations Solutions
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c d4x x2 7- =-^ h 6x x4 5- =-^ h
Page 16 questions
Page 17 questions
2, 7, 4a b c= =- =
x x2 7 4 02- + =
xa
b b ac
x
x
24
2 2
7 7 4 2 4
47 49 32
2
2
!
!
!
= - -
=- - - -
= -
^^ ^ ^ ^
hh h h h
or
or
x4
7 17= +
x10
4 2 34=-
- +
x4
7 17= -
x10
4 2 34=-
- -
5
4
104
104
xa
b b ac
x
x
x
24
2
4 4 5 6
136
34 4
2
2
!
!
!
! #
= - -
=-
- - -
=-
-
=-
-
^^ ^ ^
hh h h
5 4 6 0x x2- + + =
5, 4, 6a b c=- = =
3. Solve 7 42 112 0x x2+ - = using each of the following methods.
(You should get the same solutions each time)
Factorise and solve for x.a
x x
x x
x x
7 42 112 0
7 6 16 0
7 2 8 0
2
2
+ - =
+ - =
- + =^ ^h h
6
6
@
@
or
or
x 2 0- =
x 2=
x 8 0+ =
x 8=-
(By Null Factor Law)
11100% Quadratic Equations Solutions
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Page 17 questions
or
or
x 8=-
x 2=
x 2=
x 8=-
b Use completing the square to solve for x.
x x
x x
x
x
x
x
x
7 42 112 0
7 6 16 0
3 3 16 0
3 25 0
3 25
3 5
3 5
2
2
2 2
2
2
2 2!
!
+ - =
+ - =
+ - - =
+ - =
+ =
+ =
+ =
^
^
^
^ ^
h
h
h
h h
6 @
xa
b b ac
x
x
x
x
x
24
2 7
42 42 4 7 112
1442 1764 3136
1442 4900
1442 70
3 5
2
2
!
!
!
!
!
!
= - -
=- - -
= - +
= -
= -
=-
^^ ^hh h
x x7 42 112 02+ - =
7, 42, 112a b c= = =-
c Use the quadratic formula to solve for x.
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4. Use the discriminant to determine if these equations have two equal, unequal or no real solutions.(Hint: Make sure the equation is in standard form ax bx c 02
+ + = )
a
c
e
b
d
f
4 21 18 0x x2+ - = 3 7 2 0x x2
+ + =
3 11 0x x2- + = 2 18 16
2 16 18 0
x x
x x
2
2
- =
- - =
8 8 2x x2- =
Page 20 questions
b ac4
21 4 4 18
441 288
729
2
2
T = -
= - -
= +
=
^ ^h h
4, 21, 18a b c= = =-
0` T 2 and is a perfect square` the solutions will be unequal and rational
0` T 2 and is a perfect square` the solutions will be unequal and rational
0` T 2 and is not a perfect square` the solutions will be unequal and irrational
0` T = . The two solutions will be equalThese equal solutions will also be rational
0` T 1 ` there are no real solutions for x
0` T 2 and is a perfect square` the solutions will be unequal and rational
b ac4
49 4 3 2
25
2T = -
= -
=
^ ^h h
3, 7, 2a b c= = =
1, 3, 11a b c= =- =
b ac4
3 4 1 11
9 44
35
2
2
T = -
= - -
= -
=-
^ ^ ^h h h
400
20
b ac4
16 4 2 18
256 144
2
2
2
T = -
= - - -
= +
=
=
^ ^ ^h h h
2, 16, 18a b c= =- =-
b ac4
4 4 1 3
16 12
28
2
2
T = -
= - -
= +
=
^ ^h h
xx
x x
x x
3 4
3 4
4 3 0
2
2
= -
= -
+ - =
1, 4, 3a b c= = =-b ac4
8 4 2 8
64 64
0
2
2
T = -
= - - -
= -
=
^ ^h h
, ,
x x
a b c
2 8 8 0
2 8 8
2- + - =
=- = =-
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5. Find the value of r if rx x3 9 02- + = :
Page 21 questions
a
b
Has one solution.
Has two unequal solutions.
6. Find the values for k such that x x k k x7 6 32+ + = - has no real solutions:
, ,
rx x
a r b c
3 9 0
3 9
2- + =
= =- =
, ,
rx x
a r b c
3 9 0
3 9
2- + =
= =- =
If there is one solution then b ac4 02T = - =
If there is one solution then 4 0b ac2T 2= -
There are no real solutions when 4 0b ac2T 1= -
so r
r
r
3 4 9 0
9 36 0
41
2- - =
- =
=
^ ^ ^h h h
36
r
r
r
r
3 4 9 0
9 36 0
9
41
2 2
2
1
1
- -
-
--
^ ^ ^h h h
rx x3 9 02- + = has one solution when r
41=
The equation rx x3 9 02- + = has unequal solutions when r
411
x x k k x7 6 32+ + = - has no real solutions when k 42
Reverse the sign of the inequality when dividing by a negative number
Reverse the sign of the inequality when dividing by a negative number
x x k k x
x x k
7 6 3
4 0
2
2
+ + = -
+ + =
k
k
k
k
4 4 1 0
4 16
416
4
2 1
1
2
2
-
- -
--
^ ^ ^h h h
14 100% Quadratic Equations Solutions
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Page 22 questions
7. Use the discriminant to prove that rx rx r2 02+ + = has equal solutions for any r:
8. Prove that x kx k kx k2 2 2 3 32 2 2+ + = - never has real roots for any k:
b ac
r r r
r r
4
2 4
4 4
0
2
2
2 2
T = -
= -
= -
=
^ ^ ^h h h
since b ac42T - is always zero for rx rx r2 02+ + = it has equal solutions for any r
b ac
k k
k k
k
4
4 2 5
40
39
2
2 2
2 2
2
T
T
T
T
= -
= - -
= -
=-
^ ^ ^h h h
x kx k kx k
x kx k
2 2 2 3 3
2 5 0
2 2 2
2 2
+ + = -
- + =
2, , 5a b k c k2= =- =
Since k2 must be positive, k39 2- must be negative.
So b ac4 02T 1= - which means there are no real roots for any value of k
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9. Find p if the following equation has two unequal real solutions:
p xx2
5 42 2 2
=--
5 4p xx
p x x
p px x
px x p
2 2 2
2 2 2 5 4
2 2 10 8
2 8 2 10 0
2
2
2
2
=--
- = -
- = -
- + + - =
^ ^h h
So 2 , 8, 2 10a p b c p=- = = -
b ac
p p
p p
p p
p p
p p
4 0
8 4 2 2 10 0
64 16 80 0
16 5 4 0
5 4 0
1 4 0
2
2
2
2
2
T 2
2
2
2
2
2
= -
- - -
+ -
- +
- +
- -
^ ^
^ ^
h h
h h
6 @
5 4p p2- +
-2 -1 0 1 2 3 4 5 6
4
3
2
1
-1
-2
y
x
We can plot this to interpret the points 1p = and p 4= .
Plotting the discriminant function, p p5 42- + , we can see it is positive (above the x-axis), when p 11 or p 42 .
So p x
x25 42 2 2
=-- has two unequal and real solutions, when 1p = and p 4= .
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Page 24 questions
1. Substitute k x3= to solve for x in x x2 4 16 06 3
+ - = .
2. Make a substitution to solve for x in:
a b
c
40 6 0x x7 3 7 3 2+ + - + =^ ^h h 5 36 0x x2 24 2- - - - =^ ^h h
9 10 9 03x x- + =^ h
Let k x3=
0
k k
k k
2 4 16 0
2 4 4
2`
`
+ - =
- + =^ ^h h
or
or
or
or
or
or
or
or
or
or
or
or
or
or
k2 4 0- =
x 23` =
x7 3 4` + =-
3 9x` =
x 2 92` - =^ h
x 1` =-
y 10=
b 1=
4a =-
k 2=
x 23` =
x 1` =-
x 2` =
x 2 32 2` !- =^ ^h h
x 2 3` !- =
k 4=-
x 43=-
x7 3 10+ =
3 1x=
x 2 42- =-^ h
x 5=
y 4=-
b 9=
9a =
k 4=-
x 43= -
x 1=
x 0=
x 2 42
!- = -^ ^h h
let 7 3y x= +
y y
y y
y y
40 6 0
6 40 0
4 10 0
2
2
+ - =
- - =
+ - =^ ^h h
let a x 2 2= -^ h
a a
a a
5 36 0
9 4 0
2- - =
- + =^ ^h h
10 9 03 3x x2- + =^ ^h h
let b 3x=
b b
b b
10 9 0
9 1 0
2- + =
- - =^ ^h h
This is impossible
Since 9 33 3x x x x2 2 2= = =^ ^h h
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Area x x
x x
x x
4 3 126
4 3 126
4 3 126 0
2
2
#= - =
- =
- - =
^ h
Page 26 questions
orx 6= 21x4
=-
1. The product of two consecutive integers is 272.
Find the two numbers if they are positive.a
b Find the two numbers if they are negative.
2. A rectangle’s length is 3 less than four times the breadth. Find the dimensions of the rectangle if the area is cm126 2 :
Call the first number n
n n
n n
n n
1 272
272 0
16 17 0
2
+ =
+ - =
- + =
^
^ ^
h
h h
So for positive numbers n = 16 and n + 1 = 17
So for negative numbers n = -17 and n + 1 = -16
(the numbers are 16 and 17)
(the numbers are -17 and -16)
n n
n n
n n
1 272
272 0
16 17 0
2
+ =
+ - =
- + =
^
^ ^
h
h h
xa
b b ac2
4
2 4
3 3 4 4 126
83 2025
83 45
2
2
!
!
!
!
= - -
=- - - - -
=
=
^^ ^ ^ ^
hh h h h
Negative length doesn't make sense so breadth is 6cm and length is cm4 6 3 21# - =
Breadth = x
Length = 4x - 3
Area Breadth Length 126#= =
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Page 30 questions
3. The stage below is made up of a square and a rectangle. Find x If the total area of the stage is m191 2 .
x + 3
x + 4
x + 2 Total area = Area of square + Area of rectangle
x x x x
x x x x
x x
x x
2 2 3 4 191
4 4 7 12 191
2 11 175 0
7 2 25 0
2 2
2
+ + + + + =
+ + + + + =
+ - =
- + =
^ ^ ^ ^
^ ^
h h h h
h h
x225=-7x = or
Negative length not possible so x = 7m
4. Solve these simultaneous equations for x and y. How many times does the straight line intersect the parabola?
y x x
y x
13 1
8 1
2= + +
= +
x x x
x x
x x
13 1 8 1
5 0
5 0
2
2
+ + = +
+ =
+ =^ h
1
2
x 0= x 5=-or
Substitute x = 0 into 2
Substitute x = -5 into 2
8 1y 0
1
= +
=
^ h
y 8 5 1
39
= - +
=-
^ h
So (0,1) is one solution
So (-5,-39) is a solution
The straight line intersects the parabola twice at: ,0 1^ h and ,5 39- -^ h
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Page 31 questions
1
1
2
2
Substitute x = 2 into 2
Substitute x = 1 into 2
y 2 2 3
1
= -
=
^ h
y 2 1 3
1
= -
=-
^ h
So (2,1) is a solution
So (1,-1) is a solution
The straight line intersects the parabola twice at: ,2 1^ h and ,1 1-^ h
5. Solve these simultaneous equations for x and y. How many times does the straight line intersect the parabola?
y x x
y x
5 4 10
3 2
2- = -
+ =
y x x
y x
4 10 5
2 3
2= - +
= -
xa
b b ac2
4
2 4
12 12 4 4 8
812 16
2
2
!
!
!
= - -
=- - - -
=
^^ ^ ^ ^
hh h h h
x x x
x x
4 10 5 2 3
4 12 8 0
2
2
- + = -
- + =
x 2= x 1=or
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6. Solve these simultaneous equations for x and y. Find the coordinates of the point(s) where the graphs would intersect:
y x x
x y
3 37
23 7
2- = -
- = -
1
1
2
2
y x x
y x
3 37
7 23
2= + -
= +
x x x
x x
x x
3 37 7 23
4 60 0
10 6 0
2
2
+ - = +
- - =
- + =^ ^h h
Substitute x = 10 into 2
Substitute x = -6 into 2
y 7 10 23
93
#= +
=
7 23y 6
19
#= - +
=-
^ h
So (10,93) is one part of intersection
So (-6,-19) is one part of intersection
The graphs intersect at: ,10 93^ h and ,6 19- -^ h
x 10= x 6=-or
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Page 33 questions
1
2
7. Solve these simultaneous equations for x and y. Find the coordinates of the point(s) where the graphs would intersect:
y x x
y x
3 9 30
2 15
2= - +
=- +
x x x
x x
3 9 30 2 15
3 7 15 0
2
2
- + =- +
- + =
7
131
xa
b b ac2
4
2 3
7 4 3 15
67
2
2
!
!
!
= - -
=- - - -
= -
^^ ^ ^ ^
hh h h h
The discriminant is 01 so this can't be solved (no real solutions)
` the graphs of y x x3 9 302= - + and y x2 15=- + do not intersect each other.
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