Download More about Equations

4
NON-FOUNDATION
More about Equations
Name :
4A
4.1
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Date :
Mark :
Equations Reducible to Quadratic Equations
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Key Concepts and Formulae
The following types of equations can be reduced to quadratic equations.
(a) ax
2n
+ bx + c = 0, where a ≠ 0 and n is a positive number.
n
(b) Equations with square root signs.
Solve the following equations. (1 – 11)
1.
x − 5x + 4 = 0
4
2
Solution
By substituting x = u into the equation x − 5x + 4 = 0, we have
2
) − 5(
(
u2
(
u − 1
u = (
1
)(
4
u
) + 4 = 0
) = 0
u − 4
) or u = (
4
Q
x = u
∴
x = (
1
) or x = (
x = (
±1
) or
∴
2
)
2
2
2
x = (
4
)
±2
)
The real roots of the equation are (
−2
), (
−1
), (
1
) and (
2
).
105
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Number and Algebra
2.
x − 8x + 15 = 0
4
2
Solution
By substituting x2 = u into the equation x4 − 8x2 + 15 = 0, we have
u 2 − 8u + 15 = 0
(u − 3)(u − 5) = 0
u = 3 or u = 5
Q x2 = u
∴ x2 = 3
or
∴ x = ± 3
or
x2 = 5
x = ± 5
∴ The real roots of the equation are − 5 , − 3 ,
3.
3 and
5.
x + 28x + 27 = 0
6
3
Solution
By substituting x3 = u into the equation x6 + 28x3 + 27 = 0, we have
u 2 + 28u + 27 = 0
(u + 1)(u + 27) = 0
u = −1 or u = −27
Q x3 = u
∴ x 3 = −1
or
x 3 = −27
∴ x = −1
or
x = −3
∴ The real roots of the equation are −3 and −1.
106
4
4.
More about Equations
x + x + 20x = 0
6
4
2
Solution
x 6 + x 4 − 20x 2 = 0
x 2(x4 + x2 − 20) = 0
x 2 = 0 or x 4 + x 2 − 20 = 0
By substituting x 2 = u into the equation x 4 + x 2 − 20 = 0, we have
u 2 + u − 20 = 0
(u − 4)(u + 5) = 0
u = 4 or u = −5
Q x2 = u
∴ x2 = 4
x 2 = −5 (rejected)
or
∴ x = ±2
∴ The real roots of the equation are −2, 0 and 2.
x 2 − 3 x − 27 = 1
5.
Solution
x 2 − 3 x − 27 = 1
( x 2 − 3 x − 27 ) 2 = 12
(
x + 4
x = (
−4
Checking:
(
x 2 − 3x − 27
) = (
(
x 2 − 3x − 28
) = 0
)(
)
) = 0
x − 7
) or x = (
When x = (
1
)
7
−4
),
x 2 − 3 x − 27 =
= (
When x = (
7
),
x 2 − 3 x − 27 =
= (
∴
The real roots of the equation are (
−4
) and (
−4
(
7
1
7
−4
) − 27
) 2 − 3(
7
) − 27
)
1
(
) 2 − 3(
)
).
107
Number and Algebra
6.
x−
2 x + 20 = 2
7.
Solution
x −
x−7 x +6 = 0
Solution
x −7 x + 6 = 0
2x + 20 = 2
x −2 =
x + 6 = 7 x
2x + 20
(x + 6)2 = (7 x )2
(x − 2)2 = ( 2x + 20 )2
x 2 − 4x + 4 = 2x + 20
x 2 + 12x + 36 = 49x
x 2 − 6x − 16 = 0
x 2 − 37x + 36 = 0
(x + 2)(x − 8) = 0
(x − 1)(x − 36) = 0
x = −2
x = 1
or
x = 8
Checking:
Checking:
When x = −2,
When x = 1,
x −
2x + 20 = −2 −
2(−2) + 20
x − 7 x + 6 = 1− 7 1 + 6
= −6
= 0
≠ 2
Hence, −2 is not a root of the required
When x = 36,
x − 7 x + 6 = 36 − 7 36 + 6
equation.
= 0
When x = 8,
x −
x = 36
or
∴ The real roots of the equation are 1 and
2x + 20 = 8 −
2(8) + 20
= 2
∴ The real root of the equation is 8.
36.
Alternative solution
x = u into the equation
By substituting
x − 7 x + 6 = 0 , we have
u 2 − 7u + 6 = 0
(u − 1)(u − 6) = 0
u = 1 or u = 6
Q
x = u
∴
x = 1
or
x = 6
∴
x = 1
or
x = 36
∴ The real roots of the equation are 1 and
36.
108
4
8.
9x − 10x + 1 = 0
4
2
10.
Solution
13 − 2 x + 1 = 2 x
Solution
By substituting x 2 = u into the equation
13 − 2x + 1 = 2x
9x4 − 10x2 + 1 = 0, we have
9u 2 − 10u + 1 = 0
u =
13 − 2x = 4x 2 − 4x + 1
or u = 1
9
4x 2 − 2x − 12 = 0
Q x2 = u
∴ x2 =
13 − 2x = 2x − 1
( 13 − 2x )2 = (2x − 1)2
(9u − 1)(u − 1) = 0
1
More about Equations
2x 2 − x − 6 = 0
1
or
9
1
3
∴ x = ±
x2 = 1
x = ±1
or
(x − 2)(2x + 3) = 0
x = 2
x = −
or
3
2
Checking:
∴ The real roots of the equation are −1,
−
1
3
,
1
and 1.
3
When x = 2,
L.H.S. =
13 − 2(2) + 1
= 4
R.H.S. = 2(2)
9.
= 4
8x + 9x + 1 = 0
6
3
∴ L.H.S. = R.H.S.
Solution
Checking:
By substituting x = u into the equation
3
8x6 + 9x3 + 1 = 0, we have
8u 2 + 9u + 1 = 0
When x = −
L.H.S. =
(8u + 1)(u + 1) = 0
1
u = −
8
or
∴ x
= −
∴ x = −
13 − 2 −
3
+1
2
= 5
u = −1
3
R.H.S. = 2 − 
2
Q x3 = u
3
3
,
2
= −3
1
8
1
2
or
or
x
3
= −1
x = −1
∴ L.H.S. ≠ R.H.S.
Hence, 2 is the root of the required
equation.
∴ The real roots of the equation are −1
and −
1
.
2
109
Number and Algebra
11.
3x + 1 + 2 = 1 − 3x
Solution
3x + 1 + 2 = 1 − 3x
3x + 1 = −3x − 1
( 3x + 1)2 = (−3x − 1)2
3x + 1 = 9x 2 + 6x + 1
9x 2 + 3x = 0
1
= u , solve
x +1
2
1
− 1 = 0.
+ 2
2
2
( x + 1)
x +1
12. By substituting
2
Solution
1
By substituting
x
equation
2
+1
2
(x
we have
2
+ 1)
2
+
= u into the
1
x
2
x
2
+1
− 1 = 0,
3x 2 + x = 0
2u 2 + u − 1 = 0
x (3x + 1) = 0
x = 0
x = −
or
1
3
Checking:
When x = 0,
3(0) + 1 + 2
= 3
R.H.S. = 1 − 3(0)
∴ L.H.S. ≠ R.H.S.
Hence, 0 is not a root of the required
1
or u = −1
2
u =
Q
L.H.S. =
∴
1
x2 + 1
1
x
2
+1
When x = −
1
,
3
L.H.S. =
1
3 −  + 1 + 2
3
= 2
1
R.H.S. = 1 − 3 − 
3
= 2
∴ L.H.S. = R.H.S.
∴ The real root of the equation is 2.
= u
=
1
2
x2 = 1
or
or
1
+1
= −1
x 2 = −2 (rejected)
x = ±1
∴ The real roots of the equation are −1
and 1.
equation.
110
(2u − 1)(u + 1) = 0
4
1
1

13. By substituting x +
= u, solve 2  x + 
x
x
2
More about Equations
1

− 9  x +  + 10 = 0.
x
Solution
By substituting x +
1

= u into the equation 2  x + 
x
x
1
2
1

− 9  x +  + 10 = 0, we have
x
2u 2 + 9u + 10 = 0
(2u − 5)(u − 2) = 0
u =
5
or u = 2
2
1
= u
x
Q x +
∴ x +
1
x
=
5
2
or
x +
1
x
= 2
Multiplying both sides by x, we have
x  x +
5
1
= x

2
x
x2 + 1 =
2
x
or
x  x +
1
x
= 2x
x 2 + 1 = 2x
− 5x + 2 = 0
or
x 2 − 2x + 1 = 0
(2x − 1)(x − 2) = 0
or
(x − 1)2 = 0
2x
∴
5
or
x =
2
1
2
or x = 2
or
∴ The real roots of the equation are
x = 1 (double root)
1
, 1 (double root) and 2.
2
111
4
More about Equations
Name :
4B
4.2
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Date :
Mark :
Solving Simultaneous Equations by the Graphical Method
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Key Concepts and Formulae
Nature of solutions
The number of solutions for simultaneous equations, one linear and one quadratic,
depends on the number of points of intersections of the two graphs.
Graphs of simultaneous
equations, one linear
and one quadratic
x
x
0
0
No. of intersections
No. of solutions
y
y
y
x
0
2
1
0
Two distinct real
solutions
One real
solution
No real
solutions
The figure shows the graph of y = x for −3 ≤ x ≤ 3. Solve the following pairs of simultaneous
equations by drawing suitable straight lines on the graph. (1 – 2)
2
y
9
y = − 2x + 3
8
7
y=x2
6
5
4
y=4
3
2
1
x
−3
112
−2
−1
0
1
2
3
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4
1.
More about Equations
y = x2

y = 4
Solution
Draw the straight line (
) on the graph of y = x .
y = 4
Q The two graphs intersect at (
2
−2
,
) and (
4
∴ The solutions of the simultaneous equations are ( −2
2.
,
2
,
).
4
4
) and (
2
,
4
).
y = x2

 y = −2 x + 3
Solution
Draw the straight line y = −2x + 3 on the graph of y = x 2.
Q The two graphs intersect at (−3, 9) and (1, 1).
∴ The solutions of the simultaneous equations are (−3, 9) and (1, 1).
The figure shows the graph of y = x − x − 1 for −
2
5
7
≤ x ≤ . Solve the following pairs of
2
2
simultaneous equations by drawing suitable straight lines on the graph. (3 – 4)
8
y = x 2− x − 1
7
6
5
y=3−x
y = 3x − 4
4
3
2
1
−3
−2
−1
0
−1
x
1
2
3
4
−2
113
Number and Algebra
3.
y = x2 − x − 1

y = 3 − x
Solution
Draw the straight line y = 3 − x on the graph of y = x 2 − x − 1.
Q The two graphs intersect at (−2, 5) and (2, 1).
∴ The solutions of the simultaneous equations are (−2, 5) and (2, 1).
4.
y = x2 − x − 1

y = 3x − 4
Solution
Draw the straight line y = 3x − 4 on the graph y = x 2 − x − 1.
Q The two graphs intersect at (1, −1) and (3, 5).
∴ The solutions of the simultaneous equations are (1, −1) and (3, 5).
Write a quadratic equation which can be solved by drawing each of the following straight lines on
the graph of y = x − x. (5 – 6)
2
5.
y = 2x + 7
Solution
y = x2 − x
The simultaneous equations are 
.
y = 2 x + 7
x − x = (
∴
2
(
∴
114
x 2 − 3x − 7
2x + 7
)
) = 0
The corresponding quadratic equation is (
x 2 − 3x − 7
) = 0.
4
6.
More about Equations
3y = 6x + 2
Solution
y = x 2 − x

Q The simultaneous equations are 
.
2
y = 2x +

3
∴
x 2 − x = 2x +
x 2 − 3x −
2
3
2
= 0
3
3x 2 − 9x − 2 = 0
∴ The corresponding quadratic equation is 3x 2 − 9x − 2 = 0.
Suppose the graph of y = x + 2 is given. Find the equations of the straight lines needed to be
drawn on the given graph so as to solve the following equations. (7 – 9)
2
7.
x + 5x + 5 = 0
2
Solution
Q
x + 5x + 5 = 0
2
x + 2 = (
2
∴
8.
−5x − 3
)
The equation of the required straight line is (
y = −5x − 3
).
x − 6x − 5 = 0
2
Solution
Q x2 − 6x − 5 = 0
x2 + 2 = 6x + 7
∴ The equation of the required straight line is y = 6x + 7.
115
Number and Algebra
9.
2x − 2x + 9 = 0
2
Solution
Q 2x 2 − 2x + 9 = 0
x2 − x +
9
= 0
2
x2 + 2 = x −
5
2
∴ The equation of the required straight line is y = x − 5 .
2
Suppose the graph of y = x + 3x − 5 is given. Find the equations of the straight lines needed to
be drawn on the given graph so as to solve the following equations. (10 – 11)
2
10. 2x − 5x + 8 = 0
2
Solution
Q 2x 2 − 5x + 8 = 0
∴ x2 −
5
x + 4 = 0
2
x 2 + 3x − 5 =
11
2
x −9
∴ The equation of the required straight line is y =
11
x − 9.
2
11. 3x − 9x − 1 = 0
2
Solution
Q 3x 2 − 9x − 1 = 0
∴ x 2 − 3x −
1
= 0
3
x 2 + 3x − 5 = 6x −
14
3
∴ The equation of the required straight line is y = 6x −
116
14
3
.
4
More about Equations
Solve the following simultaneous equations graphically. (12 – 13)
12.  y = x 2 − 1

y = x + 1
Solution
y = x − 1
2
x
−3
−2
−1
0
1
2
3
y
8
3
0
−1
0
3
8
x
−1
0
1
y
0
1
2
y = x + 1
y
8
y = x 2− 1
6
4
y=x+1
2
x
−3
−2
0
−1
1
Q
The two graphs intersect at ( −1
∴
The solutions of the simultaneous equations are (
,
0
2
) and (
3
,
2
−1
,
3
0
).
) and (
2
,
3
).
117
Number and Algebra
2
13.  y = 2 x

5 x + y − 3 = 0
Solution
y = 2 x2

5 x + y − 3 = 0
y = 2 x2

 y = ( 3 − 5x
y = 2x
)
2
x
−3
−2
−1
0
1
2
3
y
18
8
2
0
2
8
18
y = (
3 − 5x
)
x
−3
−1
1
y
18
8
−2
y
y = 2x 2
y = 3 − 5x
15
10
5
x
−3
−2
−1
0
1
2
3
1 1
Q The two graphs intersect at (−3, 18) and ( , ).
2 2
∴ The solutions of the simultaneous equations are (−3, 18) and (
118
1 1
, ).
2 2
4
More about Equations
14. (a) Plot the graph of y = x − 2x + 1 for −2 ≤ x ≤ 4.
[Suggestion: Scale for x-axis is 10 divisions (1 cm) = 1 unit;
scale for y-axis is 10 divisions (1 cm) = 2 units]
2
(b) Solve the following equations by drawing suitable straight lines on the graph in (a).
(i)
x − 6x + 8 = 0
(ii) 2x + 2x − 1 = 0
2
2
Solution
(a)
y
8
6
y = 4x − 7
4
y = −3x + 3
2
y = x 2 − 2x + 1
2
x
−2
(b) (i)
−1
1
0
2
3
4
y = x 2 − 2x + 1
x 2 − 6x + 8 = 0 corresponding to the simultaneous equations 
.
y = 4x − 7
Draw the straight line y = 4x − 7 on the graph of y = x 2 − 2x + 1.
From the graph, the roots of x 2 − 6x + 8 = 0 are 2 and 4.
(ii)
y = x 2 − 2x + 1

2x + 2x − 1 = 0 corresponding to the simultaneous equations are 
.
3
y = −3x +

2
3
2
y
=
−
3
x
+
Draw the straight line
on the graph of y = x − 2x + 1.
2
2
From the graph, the roots of 2x 2 + 2x − 1 = 0 are approximately equal to −1.4 and 0.4.
119
4
NON-FOUNDATION
More about Equations
Name :
4C
Date :
Mark :
4.3 Solving Simultaneous Equations by the Algebraic Method
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Key Concepts and Formulae
 y = ax 2 + bx + c
For the simultaneous equations 
, the number of solutions
 y = mx + n
2
depends on the discriminant of ax + (b − m)x + (c − n) = 0.
Discriminant of
ax2 + (b − m)x + (c − n) = 0
No. of solutions of
y = ax 2 + bx + c

y = mx + n
∆ > 0
∆ = 0
∆ < 0
Two distinct
real solutions
One real
solution
No real
solutions
Solve the following simultaneous equations using the algebraic method. (1 – 5)
1.
y = x2 − 2

y = 3x − 4
Solution
y = x2 − 2

y = 3x − 4
......(1)
......(2)
By substituting (2) into (1), we have
3x − 4 = (
(
x − 1
x = (
1
x = (
) or
By substituting x = (
y = 3(
120
)= 0
x − 2
)(
1
)
)= 0
x 2 − 3x + 2
(
x2 − 2
1
) − (
2
)
) into (2), we have
4
) = (
−1
)
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4
By substituting x = (
y = 3(
∴
2.
2
2
) − (
) into (2), we have
4
) = (
2
)
The solutions of the simultaneous equations are (
y = 2 x − 5
 2
2
 x + y = 85
3.
Solution
y = 2x − 5
 2
2
x + y = 85
More about Equations
, −1
1
) and (
2
,
2
).
− y 2 = x 2 + x − 2

y = x + 2
Solution
......(1)
......(2)
By substituting (1) into (2), we have
x 2 + (2x − 5)2 = 85
x 2 + 4x 2 − 20x + 25 = 85
5x 2 − 20x − 60 = 0
x 2 − 4x − 12 = 0
(x + 2)(x − 6) = 0
x = −2 or x = 6
− y 2 = x 2 + x − 2

y = x + 2
By substituting x = 6 into (1), we have
y = 2(6) − 5 = 7
∴ The solutions of the simultaneous
equations are (−2, −9) and (6, 7).
......(2)
By substituting (2) into (1), we have
−(x + 2)2 = x 2 + x − 2
− x 2 − 4x − 4 = x 2 + x − 2
2x 2 + 5x + 2 = 0
(2x + 1)(x + 2) = 0
x = −
1
2
or x = −2
By substituting x = −
By substituting x = −2 into (1), we have
y = 2(−2) − 5 = −9
......(1)
y = −
1
2
+2 =
1
into (2), we have
2
3
2
By substituting x = −2 into (2), we have
y = −2 + 2 = 0
∴ The solutions of the simultaneous
 1 3
equations are − ,
and (−2, 0).
 2 2
121
Number and Algebra
4.
x − y = 1

2
2
6 xy − 4 x + y = −4
5.
Solution
x − y = 1

2
2
6xy − 4x + y = −4
...… (1)
...… (2)
From (1), we have
y = x − 1
...… (3)
By substituting (3) into (2), we have
6x (x − 1) − 4x 2 + (x − 1)2 = −4
3x 2 − 8x + 5 = 0
(3x − 5)(x − 1) = 0
x =
5
3
or x = 1
5
By substituting x =
into (3), we have
3
y =
5
2
− 1 =
3
3
By substituting x = 1 into (3), we have
y = 1 −1 = 0
∴ The solutions of the simultaneous
 5 2
,
equations are
and (1, 0).
 3 3
y2 − 2 x2 = 2 y − 8

y − 2
 2 + 3 − x = 0
Solution
y 2 − 2x 2 = 2y − 8

y − 2
+3− x = 0

 2
...… (1)
...… (2)
From (2), we have
y − 2
= x −3
2
y = 2x − 4
...… (3)
By substituting (3) into (1), we have
(2x − 4)2 − 2x 2 = 2(2x − 4) − 8
4x 2 − 16x + 16 − 2x 2 = 4x − 16
2x 2 − 20x + 32 = 0
x 2 − 10x + 16 = 0
(x − 2)(x − 8) = 0
x = 2 or x = 8
By substituting x = 2 into (3), we have
y = 2(2) − 4 = 0
By substituting x = 8 into (3), we have
y = 2(8) − 4 = 12
∴ The solutions of the simultaneous
equations are (2, 0) and (8, 12).
122
4
More about Equations
Find the number of solutions of the following simultaneous equations. (6 – 8)
6.
y = 1 − 3x

2
y = 4x + 5
Solution
y = 1 − 3x

2
y = 4x + 5
...… (1)
...… (2)
By substituting (1) into (2), we have
1 − 3x = (
4x 2 + 3x + 4
(
4x 2 + 5
) = 0
7.
) − 4(
2
3
)(
4
4
Q
∆(
∴
(
∴
The simultaneous equations have (
<
) = 0.
4x 2 + 3x + 4
Consider the discriminant of (
∆ = (
)
) = (
−55
)
)0
4x 2 + 3x + 4
) = 0 has (
no
no
) real roots.
) real solutions.
2 x 2 + y + 4 x − 1 = 0
 2
 x + y = 10(1 − x)
Solution
2x 2 + y + 4x − 1 = 0

x 2 + y = 10(1 − x )
...… (1)
...… (2)
From (1), we have
y = −2x 2 − 4x + 1
...… (3)
By substituting (3) into (2), we have
x 2 −2x 2 − 4x + 1 = 10(1 − x)
x 2 − 6x + 9 = 0
Consider the discriminant of x 2 − 6x + 9 = 0.
∆ = (−6)2 − 4(1)(9) = 0
∴ x 2 − 6x + 9 = 0 has a double real root.
∴ The simultaneous equations have one real solution.
123
Number and Algebra
8.
y2 − 6 = 0

2 x − y = 1
Solution
y 2 − 6 = 0

2x − y = 1
...… (1)
...… (2)
From (2), we have
y = 2x − 1
...… (3)
By substituting (3) into (1), we have
(2x − 1)2 − 6 = 0
4x 2 − 4x + 1 − 6 = 0
4x 2 − 4x − 5 = 0
Consider the discriminant of 4x 2 − 4x − 5 = 0.
∆ = (−4)2 − 4(4)(−5) = 96 > 0
∴ 4x 2 − 4x − 5 = 0 has two distinct real roots.
∴ The simultaneous equations have two distinct real solutions.
9.
y = k − x2
Given that the simultaneous equations 
have only one solution, find the value of k.
 y = −10 x
Solution
y = k − x2

 y = −10 x
......(1)
......(2)
By substituting (1) into (2), we have
( k − x2 ) = (
) = 0
)
(
x 2 − 10x − k
Q
The simultaneous equations have only one solution.
∴
(3) has ( one ) real root.
∴
∆ ( = ) 0
(−10)2 − 4(1)(−k) = 0
100 + 4k = 0
k = −25
124
−10x
......(3)
4
10. Given that the simultaneous equations
 x 2 − y 2 = 5( x − k )
have two distinct

y = 2 x − 3
real solutions, find the range of possible
values of k.
 x 2 − y 2 = 5(x − k )

y = 2x − 3
...… (1)
x 2 − (2x − 3)2 = 5(x − k)
x − 4x + 12x − 9 = 5x − 5k
3x
k − y = 2 x + 1

2
y = x + 4x − 7
(b) Hence, solve the simultaneous
equations.
...… (2)
By substituting (2) into (1), we have
2
2
11. It is given that the following
simultaneous equations have only one
real solution.
(a) Find the value of k.
Solution
2
More about Equations
− 7x + (9 − 5k) = 0
...… (3)
Q The simultaneous equations have two
distinct real solutions.
Solution
(a) k − y = 2x + 1
...… (1)

2
y = x + 4x − 7
...… (2)
From (1), we have
y = k − 2x − 1
...… (3)
By substituting (3) into (2), we have
∴ (3) has two distinct real roots.
k − 2x − 1 = x2 + 4x − 7
∴ ∆ > 0
x 2 + 6x − 6 − k = 0
i.e. (−7) − 4(3)(9 − 5k) > 0
Q The simultaneous equations have
2
49 − 108 + 60k > 0
60k > 59
k >
59
60
only one real solution.
∴ (4) has only one real root.
∆ = 0
(6)2 − 4(1)(−6 − k) = 0
∴ The range of possible values of k is
k >
59
60
...… (4)
4k = −60
k = −15
.
(b) From (4), we have
x 2 + 6x − 6 − (−15) = 0
x 2 + 6x + 9 = 0
(x + 3)2 = 0
x = −3 (double root)
By substituting x = −3 into (3), we have
y = −15 − 2(−3) − 1 = −10
∴ The solution of the simultaneous
equations is (−3, −10).
125
Number and Algebra
12. It is given that the following simultaneous equations have real solutions.
3 x 2 − y 2 = k

x − y + 4 = 0
(a) Find the range of possible values of k.
(b) For the minimum value of k, solve the simultaneous equations. (Leave the answers in surd
form.)
Solution
(a) 3x 2 − y 2 = k
...… (1)

x − y + 4 = 0
...… (2)
From (2), we have
y = x + 4
...… (3)
By substituting (3) into (1), we have
3x 2 − (x + 4)2 = k
3x 2 − x 2 − 8x − 16 = k
2x 2 − 8x − 16 − k = 0
...… (4)
Q The simultaneous equations have real solutions.
∴ (4) has real roots.
∆ ≥ 0
(−8) − 4(2)(−16 − k) ≥ 0
2
8k ≥ −192
k ≥ −24
∴ The range of possible values of k is k ≥ −24.
(b) From (a), the minimum value of k is −24.
From (4), we have
2x 2 − 8x − 16 − (−24) = 0
2x 2 − 8x + 8 = 0
x 2 − 4x + 4 = 0
(x − 2)2 = 0
x = 2 (double root)
By substituting x = 2 into (3), we have
y = 2 + 4 = 6
∴ The solution of the simultaneous equations is (2, 6).
126
4
More about Equations
13. The sum of two integers is 5 and the sum of their squares is 37. Find the two integers.
Solution
Let x and y be the two integers.
x + y = (

( x 2 + y 2
5
)
) = 37
...… (1)
...… (2)
From (1), we have
y = (
5 − x
)
…... (3)
By substituting (3) into (2), we have
x 2 + (5 − x )2 = 37
x 2 + 25 − 10x + x 2 = 37
2x 2 − 10x − 12 = 0
x 2 − 5x − 6 = 0
(x − 6)(x + 1) = 0
x = (
6
)
or x = (
By substituting x = (
y = (
5 − 6
By substituting x = (
y = (
∴
5 − (−1)
6
−1
) into (3), we have
) = (
−1
−1
)
) into (3), we have
) = (
The two integers are (
)
−1
6
)
) and (
6
).
127
Number and Algebra
14. The length of the base of a parallelogram is longer than its height by 3 cm. If the area of the
2
parallelogram is 18 cm , find the length of the base and the height of the parallelogram.
Solution
Let x cm and y cm be the length of base and height of the parallelogram respectively.
x = y + 3

 xy = 18
...… (1)
...… (2)
From (1), we have
y = x − 3
...… (3)
By substituting (3) into (2), we have
x (x − 3) = 18
x − 3x − 18 = 0
2
(x − 6)(x + 3) = 0
...… (4)
x = 6 or x = −3 (rejected)
By substituting x = 6 into (3), we have
y = 6 − 3 = 3
∴ The length of the base and the height of the parallelogram are 6 cm and 3 cm respectively.
15. Two squares are formed by two copper wires. If the difference in the lengths of the two wires is
2
8 cm and the sum of the areas of the two squares is 100 cm , find the lengths of the two wires.
Solution
Let x cm and y cm be the lengths of the two wires with y > x.
y = x + 8

 x  2  y  2
  +   = 100
4
 4
128
...… (1)
...… (2)
4
More about Equations
Solution
By substituting (1) into (2), we have
 x
 4
2
x + 8
+ 
 4 
2
= 100
x 2 + x 2 + 16x + 64 = 1600
x 2 + 8x − 768 = 0
(x − 24)(x + 32) = 0
x = 24 or x = −32 (rejected)
By substituting x = 24 into (1), we have
y = 24 + 8 = 32
∴ The lengths of the two wires are 24 cm and 30 cm.
16. The product of the two digits of a positive two-digit number is greater than the sum of its two
digits by 19, and its tens digit is smaller than its unit digit by 1, find the number.
Solution
Let x and y be the unit digit and the tens digit of the number respectively.
 xy − (x + y ) = 19

y = x − 1
…... (1)
...… (2)
By substituting (2) into (1), we have
x (x − 1) − [x + (x − 1)] − 19 = 0
x 2 − 3x − 18 = 0
(x − 6)(x + 3) = 0
x = 6 or x = −3 (rejected)
By substituting x = 6 into (2), we have
y = 6 − 1 = 5
∴ The number is 56.
129
4
ENRICHMENT
More about Equations
Name :
4D
4.4
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Date :
Mark :
Solving Cubic or Higher Degree Equations by
the Algebraic Method
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In each of the following: (1 – 3)
(a) Show that the expression in the bracket is a factor of f (x).
(b) Factorize f (x) completely.
(c) Solve the equation f (x) = 0.
1.
f (x) = x − 13x + 12
[x − 1]
3
(a) Q f (1) = (1)3 − 13(1) + 12
= 0
∴ x − 1 is a factor of f (x).
(b) By long division,
x
x − 1
)x
x
− 12
2
+
3
+ 0x − 13x + 12
− x2
3
x
2
2
x − 13x
2
x −
x
−12x + 12
−12x + 12
∴ x 3 − 13x + 12 = (x − 1)(x 2 + x − 12)
= (x − 1)(x − 3)(x + 4)
(c) Q
∴
f (x) = 0
( x − 1 )( x − 3 )( x + 4 ) = 0
( x − 1 ) = 0
x = (
130
or
1
)
or
( x − 3 ) = 0
x = ( 3
or
)
or
( x + 4 ) = 0
x = ( −4 )
○
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4
2.
f (x) = x + 8x + 20x + 16
3
2
[x + 2]
3.
More about Equations
f (x) = x − 4x − 17x + 60
3
2
[x − 3]
Solution
Solution
(a) Q
(a) Q f (3) = (3)3 − 4(3)2 − 17(3) + 60
f (−2)
= (−2)3 + 8(−2)2 + 20(−2) + 16
= 0
= 0
∴ (x − 3) is a factor of f (x).
∴ (x + 2) is a factor of f (x).
(b) By long division,
x
x + 2
)x
x
(b) By long division,
+ 8
2
+ 6x
3
+ 8x 2 + 20x + 16
+ 2x 2
3
x
x − 3
)x
x
2
−
3
− 4x 2 − 17x + 60
− 3x 2
3
x
− 20
2
−x − 17x
2
−x + 3 x
6 x + 20x
2
6 x + 12x
2
−20 x + 60
−20 x + 60
8 x + 16
8 x + 16
∴
x 3 + 8x 2 + 20x + 16
= (x + 2)(x
2
+ 6x + 8)
= (x + 2)(x + 2)(x + 4)
= (x + 2)2 (x + 4)
(c) Q f (x) = 0
∴ (x + 2) (x + 4) = 0
(x + 2) = 0
x 3 − 4x 2 − 17x + 60
= (x − 3)(x 2 − x − 20)
= (x − 3)(x + 4)(x − 5)
(c) Q f (x) = 0
∴ (x − 3)(x + 4)(x − 5) = 0
2
2
∴
or (x + 4) = 0
x = −2 (double root) or x = −4
x − 3 = 0 or x + 4 = 0
x = 1 or
or x − 5 = 0
x = −4 or
x = 5
131
Number and Algebra
In each of the following: (4 – 5)
(a) Show that x + 1 and x − 3 are factors of f(x).
(b) Factorize f (x) completely.
(c) Solve the equation f (x) = 0.
4.
f (x) = x − 3x − 7x + 15x + 18
4
3
2
Solution
(a) Q f (−1) = (−1)4 − 3(−1)3 − 7(−1)2 + 15(−1) + 18
= 0
Q f (3) = (3)4 − 3(3)3 − 7(3)2 + 15(3) + 18
= 0
∴ x + 1 and x − 3 are factors of f (x).
(b) (x + 1)(x − 3) = x 2 − 2x − 3
By long division,
x
x
2
− 2x − 3
)x
x
2
−
4
− 3x 3 − 7x 2 + 15x + 18
− 2x 3 − 3x 2
4
x
− 6
3
2
−x − 4x + 15x
3
2
−x + 2x + 3x
2
−6x + 12x + 18
2
−6x + 12x + 18
∴ x 4 − 3x 3 − 7x 2 + 15x + 18 = (x + 1)(x − 3)(x 2 − x − 6)
= (x + 1)(x − 3)(x + 2)(x − 3)
= (x + 1)(x + 2)(x − 3)2
(c) Q f (x) = 0
∴ (x + 1)(x + 2)(x − 3)2 = 0
x + 1 = 0
x = −1
132
or
or
x + 2 = 0
x = −2
or
or
(x − 3)2 = 0
x = 3 (double root)
4
5.
More about Equations
f (x) = x + 2x − 8x − 18x − 9
4
3
2
Solution
(a) Q f (−1) = (−1)4 + 2(−1)3 − 8(−1)2 − 18(−1) − 9
= 0
Q f (3) = (3)4 + 2(3)3 − 8(3)2 − 18(3) − 9
= 0
∴ x + 1 and x − 3 are factors of f (x).
(b) (x + 1)(x − 3) = x 2 − 2x − 3
By long division,
x
x
2
− 2x − 3
)x
x
2
+ 4x
4
+ 2x − 8x 2 − 18x − 9
− 2x 3 − 3x 2
4
+ 3
3
3
2
4x − 5x − 18x
3
2
4x − 8x − 12x
2
3x −
2
3x −
6x − 9
6x − 9
∴ x 4 + 2x 3 − 8x 2 − 18x − 9 = (x + 1)(x − 3)(x 2 + 4x + 3)
= (x + 1)(x − 3)(x + 1)(x + 3)
= (x + 1)2 (x + 3)(x − 3)
(c) Q f (x) = 0
∴ (x + 1)2(x + 3)(x − 3) = 0
(x + 1)2 = 0
x = −1 (double root)
or
or
x + 3 = 0
x = −3
or
x − 3 = 0
or
x = 3
133
Number and Algebra
In each of the following: (6 – 8)
(a) Factorize f (x) completely.
(b) Solve f (x) = 0.
6.
f (x) = x − 3x − 22x + 24
3
2
Solution
(a) Q
∴
f (1) = (
1
) − 3(
= (
0
)
x − 1
(
3
1
) − 22(
2
1
) + 24
) is a factor of f (x).
By long division,
x
x − 1
)x
x
2
− 2x
3
− 3x − 22x
− x2
3
− 24
+ 24
2
2
−2x − 22x
2
−2x + 2x
+ 24
+ 24
−24x
−24x
∴
x − 3x − 22x + 24 = (
3
2
x + 1
)(
x 2 − 2x − 24
= ( x + 1 )( x + 4 )( x − 6 )
(b) Q f (x) = 0
∴ (x − 1)(x + 4)(x − 6) = 0
134
x − 1 = 0
or
x = 1
or
x + 4 = 0
x = −4
or
x − 6 = 0
or
x = 6
)
4
7.
More about Equations
f (x) = x − 2x − 13x − 10
3
2
Solution
(a) f (1) = (1)3 − 2(1)2 − 13(1) − 10
= −24
≠ 0
f (−1) = (−1)3 − 2(−1)2 − 13(−1) − 10
= 0
∴ x + 1 is a factor of f (x).
By long division,
x
x + 1
)x
x
2
− 3x
3
− 2x − 13x − 10
+ x2
3
− 10
2
2
−3x − 13x
2
−3x − 3x
−1 0 x − 10
−1 0 x − 10
∴ x 3 − 2x 2 − 13x − 10 = (x + 1)(x 2 − 3x − 10)
= (x + 1)(x + 2)(x − 5)
(b) Q f (x) = 0
∴ (x + 1)(x + 2)(x − 5) = 0
x + 1 = 0
x = −1
or
or
x + 2 = 0
x = −2
or
x − 5 = 0
or
x = 5
135
Number and Algebra
8.
f (x) = x − 7x + 11x + 7x − 12
4
3
2
Solution
(a) f (1) = (1)4 − 7(1)3 + 11(1)2 + 7(1) − 12
= 0
∴ x − 1 is a factor of f (x).
f (−1) = (−1)4 − 7(−1)3 + 11(−1)2 + 7(−1) − 12
= 0
∴ x + 1 is a factor of f (x).
(x − 1)(x + 1) = x 2 − 1
By long division,
x
x
2
− 1
)x
x
2
− 7x
4
− 7x 3 + 11x 2 + 7x − 12
2
+ 0x 3 −
x
4
+ 12
3
2
−7x + 12x + 7x
3
2
−7x + 0x + 7x
2
12x + 0x − 12
2
12x + 0x − 12
∴ x 4 − 7x 3 + 11x 2 + 7x − 12 = (x − 1)(x + 1)(x 2 − 7x + 12)
= (x − 1)(x + 1)(x − 3)(x − 4)
(b) Q f (x) = 0
∴ (x − 1)(x + 1)(x − 3)(x − 4) = 0
136
x − 1 = 0
or
x = 1
or
x + 1 = 0
x = −1
or
x − 3 = 0
or
x − 4 = 0
or
x = 3
or
x = 4
4
9.
More about Equations
Solve the cubic equations 8(x + 2) − 27(x + 1) = 0
3
3
Solution
8(x + 2)3 − 27(x + 1)3 = 0
[2(x + 2)]3 − [3(x + 1)]3 = 0
[2(x + 2) − 3(x + 1)] {[2(x + 2)]2 + (2)(x + 2)(3)(x + 1) + [3(x + 1)]2} = 0
(2x + 4 − 3x − 3)[(2x2 + 8x + 8) + (6x2 + 18x + 12) + (3x2 + 6x + 3)] = 0
(−x + 1)(11x 2 + 32x + 23) = 0
∴ −x + 1 = 0
or
x = 1
or
11x 2 + 32x + 23 = 0
x =
−32 ±
2
− 4(11)(23)
2(11)
32
=
−32 ± 2 3
22
=
−16 ±
11
3
10. Solve the following equations by regrouping terms.
x + x + x + 3x − 6 = 0
4
3
2
4
2
3
(Hint: x + x + x + 3x − 6 = (x + x − 6) + (x + 3x))
4
3
2
Solution
x4 + x3 + x2 + 3x − 6 = 0
(x4 + x2 − 6) + (x3 + 3x) = 0
(x2 + 3)(x2 − 2) + x(x2 + 3) = 0
(x2 + x − 2)(x2 + 3) = 0
(x − 1)(x + 2)(x 2 + 3) = 0
∴ x − 1 = 0
or
x = 1
or
x + 2 = 0
x = −2
or
or
x2 + 3 = 0
x 2 = −3 (rejected)
137
4
More about Equations
Name :
4E
Date :
Mark :
Multiple Choice Questions
1.
Solve x 4 + 5 x 2 − 36 = 0 .
5.
A. x = −3 or 3
B.
x = −3, −2, 2 or 3
C.
x = −2 or 2
D. x = −9 or 4
2.
C
4
37
Solve 4 − 2 + 9 = 0.
x
x
A. x = −2 or 2
B.
C.
D.
3.
4.
1 1
or 2
3 3
1 1
or 3
x = −3, − ,
2 2
1
x = − or 2
2
6.
x = −2, − ,
B.
−1, 3
C.
D.
− 3, 3
0, 8
3
x2 + 1
A
7.
+ 5 = 0 by
substitution, which of the following is
the MOST suitable substitution of u?
A.
B.
138
x2 + 1 = u
1
x2 + 1
= u
C.
x2 + 1 = u
D.
x2 = u
x 2 + 4x − 5 = 0
B.
x 2 + 5x − 6 = 0
C.
x 2 + 5x − 5 = 0
D.
x 2 + 3x − 4 = 0
B
D
Find the quaduatic equation which can
be solved by drawing the straight line
3
x on the graph of
2
y = 3 x 2 − 10 x + 11.
B
To solve the equation
2
−
2
x +1
A.
y = 4−
Solve x − 2 x + 1 − 2 = 0.
A. 8
Find the quadratic equation which can
be solved by drawing the straight line
x − y − 1 = 0 on the graph of
y = x 2 + 4 x − 5.
A.
6 x 2 − 23 x + 30 = 0
B.
3x 2 − 2 x + 3 = 0
C.
D.
3 x 2 − 10 x + 11 = 0
6 x 2 − 17 x + 14 = 0
D
Find the equation of the straight line that
should be drawn on the graph of
y = 2 x 2 + 3 in order to solve
4 x 2 + 15 x + 12 = 0.
15
x−3
2
A.
y = −
B.
y = −15 x − 9
C.
y =
15
x+6
2
D.
y =
15
x+3
2
A
4
8.
9.
Find the equation of the straight line that
should be drawn on the graph of
y = 2 x 2 − 4 x + 5 to solve
2 x 2 + 3 x + 1 = 0.
A.
y = −7 x + 4
B.
y = −x + 6
C.
y = 7x − 4
D.
y = x−6
A
A. −2.5, −0.5
B.
−3, −1
C.
−3, 0
D. −2, 0
C
11. The figure shows the graphs of
y = x 2 + 2 x − 1 and y = x − k.
Which of the following is a possible
value of k?
The following figure shows the graphs
of y = ax 2 + bx + c and y = mx + d .
Which of the following MUST be true?
y
More about Equations
y = x 2 + 2x − 1
y
y = ax 2 + bx + c
x
0
x
0
y=x−k
y = mx + d
A.
b 2 = 4ac
B.
(b − m) 2 = 4a (c − d )
C.
(b − m) 2 > 4a (c − d )
D.
b 2 < 4ac
A. 2
B.
C.
D. −1
0
1
A
12. The figure shows the graphs of
y = ax 2 + bx + c and y = mx + d .
Which of the following MUST be true?
y
B
y = ax 2 + bx + c
10. The figure shows the graphs of
y = mx + d
y = x 2 + px + q and y = mx + c.
Find the roots of
x 2 + ( p − m) x + (q − c) = 0.
x
0
y
y = x 2 + px + q
y = mx + c
−2
−3
4a (c − d ) > (b − m) 2
The quadratic equation
ax 2 + bx + c = 0 has two real
roots.
III. The quadratic equation
ax 2 + (b − m) x + (c − d ) = 0 has
two real roots.
I.
II.
−1
x
0
139
Number and Algebra
15. The figure shows the graphs of
y = x 2 − ax + 8 and y = 2 x + b. Find
the values of a and b.
A. I only
B.
II only
C.
I and II only
y
D. II and III only
D
y = x − ax + 8
2
13. The figure shows the graphs of
2
y = x − x − k and y = kx − 2. Which of
the following is a possible value of k?
−3
y = x 2− x − k
y
x
−1 0
y = 2x + b
A. a = 6, b = −5
y = kx − 2
x
0
3
B.
C.
D. 2
0
a = −6, b = 5
C.
a = 6, b = 5
D. a = −6, b = −5
D
14. The figure shows the graphs of
y = ax + bx + c and y = k. Find the
range of possible values of k.
y = x 2 + 4x + 7
.

 y = −4 x
A. (−1, 4), (−7, 28)
2
B.
( − 7 , 4 4 ), ( 7 , −4 7 )
D. No real solutions
y=k
y = ax 2 + bx + c
x
140
4 ac − b 2
4a
4 ac − b 2
4a
B.
k =
C.
4 ac − b 2
k <
4a
D.
k ≥
4 ac − b 2
4a
A
17. Which of the following simultaneous
equations can be used to solve the
quadratic equation 2 x 2 − 7 x + 9 = 0 ?
0
k >
(0, 0), (−7, 28)
C.
y
A.
B
16. Solve the simultaneous equations
−2
A. 3
B.
C
A.
y = 2 x 2 − 5x − 1

 y = −2 x + 10
B.
x + y − 1 = 0
 2
2 x − 4 x + 8 = y
C.
y = 9 − 6x

2
y = 2 x + x
D.
 y = x2

2
7 x − y − 9 = 0
D
4
18. Find the number of real solutions of the
simultaneous equations
21. Find the minimum value of k so that the
simultaneous equations
y = 2 x + 1
have real solutions.
 2
2
4 x − 2 y = − k
y = 5 − 3x

 2
51
.
2
 x + y + 20 x − 2 = 0
A. 0
A. 0
B.
−1
B.
1
C.
−2
C.
2
D. 1
D. Cannot be determined
More about Equations
C
22. Factorize x 3 + 5 x 2 − x − 5.
A.
( x − 1) 2 ( x + 5)
kx 2 − y 2 + 1 = 0
have two
equations 
y = 2 − x
B.
C.
( x − 1)( x + 1)( x + 5)
( x + 1)( x + 5) 2
real solutions, find the range of possible
values of k.
D.
( x − 1)( x + 1)( x − 5)
19. Given that the simultaneous
1
3
A.
k ≥
B.
k > −
1
3
C.
k < −3
D.
1
k ≤ −
4
B
20. Given that the simultaneous equations
C
B
23. Factorize 27a 3 − 8(a − b) 3.
A.
B.
C.
(a + 2b)(7a 2 − 2ab + 4b 2 )
(a − 2b)(7a 2 − 2ab + 4b 2 )
(a + 2b)(19a 2 − 14ab + 4b 2 )
D.
(a − 2b)(19a 2 − 14ab + 4b 2 )
C
24. Given that x − 2 is a factor of
f ( x) = x 4 − 3 x 3 − 8 x 2 + 12 x + 16,
solve f ( x) = 0.
A. x = −2, −1, 2 or 4
 y = x 2 + ax − 7
have two real

y = 2 x − b
B.
x = −2, 1, 2 or 4
C.
solutions (−1, y1) and (4, y2). Find the
values of a and b.
x = −4, −2, 1 or 2
D. x = 1, 2, 4 or 8
A
25. Solve x 4 + x 3 − 3 x 2 − x + 2 = 0.
A.
a = −1, b = 3
B.
a = 3, b = 0
A. x = −2, −1 or 1
C.
a = 2, b = 3
B.
x = −1, 0, 1 or 2
D.
a = 0, b = 2
C.
x = 1 or 2
A
D. x = −2, −1, 1 or 2
A
141