Shatin Pui Ying College
F.4 Mathematics
Revision Exercise Ch.4
1.
##
(10/12)
Solve x – 20x + 64 = 0.
4
2
By substituting x 2  u into the equation x 4  20 x 2  64  0 , we have
u 2  20u  64  0
(u  4)(u  16)  0
u  4 or u  16
2
∵ x u
∴
∴
2.
##
x2  4
or x 2  16
x  2 or
x  4
The real roots of the equation are 4, 2, 2 and 4.
Solve x6 + 28x3 + 27 = 0.
By substituting x 3  u into the equation x 6  28 x 3  27  0 , we have
u 2  28u  27  0
(u  1)(u  27)  0
u  1 or u  27
3
∵ x u
x3  27
∴ x 3  1 or
x  1 or
x  3
∴ The real roots of the equation are 1 and 3.
3.
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Solve
x  2  x  4 .
x  2  x  4
x2  x4
( x  2 ) 2  ( x  4) 2
x  2  x 2  8 x  16
x 2  9 x  18  0
( x  3)( x  6)  0
x  3 or x  6
Checking: When x = 3,
x 2  x  32 3
 2
 4
Hence, 3 is not a root of x  2  x  4 .
x2  x  62 6
 4
The real root of the equation is 6.
When x = 6,
∴
1
4.
The figure shows the graph of y = x2 for  3  x  3 .
Solve the following quadratic equations graphically by drawing
suitable straight lines on the graph.
(a) x2 + x – 2 = 0
(b) x2 – 4x + 3 = 0
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(a)
x2  x  2  0
∴
x2   x  2
The corresponding simultaneous equations are
 y  x2
.

 y  x  2
Draw the straight line y   x  2 on the graph of y  x 2 .
From the graphs, the roots of x 2  x  2  0 are  2 and 1.
(b) x 2  4 x  3  0
x2  4x  3
∴
 y  x2
The corresponding simultaneous equations are 
.
 y  4x  3
Draw the straight line y  4 x  3 on the graph of y  x 2 .
From the graphs, the roots of x 2  4 x  3  0 are 1 and 3.
5.
Solve the following simultaneous equations.
 y  2 x 2  3x  4

2 x  y  1  0
2
##
 y  2 x 2  3x  4 ...... (1)

...... (2)
2 x  y  1  0
From (2), we have
y  2x  1
...... (3)
By substituting (3) into (1), we have
2 x  1  2 x 2  3x  4
2x2  5x  3  0
(2 x  1)( x  3)  0
1
or x  3
2
1
By substituting x   into (3), we have
2
x
 1
y  2    1  2
 2
By substituting x  3 into (3), we have
y  2(3)  1  5
∴
 1

The solutions of the simultaneous equations are   ,2  and (3, 5).
 2

6.
Solve the following simultaneous equations.
2 x 2  xy  2

x  2 y  8
##
2 x 2  xy  2 ...... (1)

...... (2)
x  2 y  8
From (2), we have y 
8 x
...... (3)
2
By substituting (3) into (1), we have
8 x
2 x 2  x
2
 2 
4 x 2  x(8  x)  4
4 x2  8x  x2  4
5x2  8x  4  0
( x  2)(5 x  2)  0
2
5
By substituting x  2 into (3), we have
x  2 or x  
y
82
3
2
By substituting x  
2
into (3), we have
5
 2
8   
 5   21
y
2
5
3
∴
 2 21 
The solutions of the simultaneous equations are (2, 3) and   ,  .
 5 5
7.
 y  2 x 2  3x  k
It is known that the simultaneous equations 
have only one solution.
4 x  y  2
(a) Find the value of k.
(b) Solve the simultaneous equations.
##
 y  2 x 2  3x  k
(a) 
4 x  y  2
...... (1)
...... (2)
From (2), we have y  4 x  2
...... (3)
By substituting (3) into (1), we have
4 x  2  2 x 2  3x  k
2 x 2  x  (k  2)  0 ...... (4)
∵
∴
The simultaneous equations have only one solution.
(4) has only one real root.
0
∴
(1) 2  4(2)( k  2)  0
1  8k  16  0
 8k  17
17
k
8
(b) By substituting k 
17
into (4), we have
8
 17

2x2  x    2   0
8

1
2x2  x   0
8
2
16 x  8 x  1  0
(4 x  1) 2  0
1
x
4
By substituting x 
1
into (3), we have
4
1
y  4   2  3
4
1 
The solution of the simultaneous equations is  , 3  .
4 
3
2
Let f(x) = x – 3x – 10x + 24.
∴
8.
(a) Show that x – 2 is a factor of f(x).
(b) Factorize f(x) completely.
(c) Hence, solve f(x) = 0.
4
##
(a) ∵
∴
f (2)  23  3(2) 2  10(2)  24
 8  12  20  24
0
x – 2 is a factor of f(x).
(b) By long division,
x 2  x  12
x  2 x 3  3x 2  10 x  24
x3  2 x 2
 x 2  10 x
 x2  2x
 12 x  24
 12 x  24
∴
(c) ∵
∴
x3  3x 2  10 x  24  ( x  2)( x 2  x  12)
 ( x  2)( x  3)( x  4)
f ( x)  0
( x  2)( x  3)( x  4)  0
x  2  0 or x  3  0
x  2 or
9.
or
x   3 or
x4  0
x4
The figure shows the graphs of y = x2 + bx + c and y = mx + k.
(a) Solve the quadratic equation x2 + (b – m)x + (c – k) = 0 graphically.
(b) Find the values of b, m, c and k of the equation in (a).
(c) If the straight line 3x + 2y = 4 is added to the graph of y = x2 + bx + c, which quadratic equation can be
solved graphically?
##
(a)
x 2  (b  m) x  (c  k )  0
x 2  bx  c  mx  k
∴
 y  x 2  bx  c
The corresponding simultaneous equations are 
.
 y  mx  k
From the graphs, the roots of x2 + (b – m)x + (c – k) = 0 are 1 and 5.
(b) ∵
y = x2 + bx + c passes through the two points (1, 3) and (5, 5).
5
∴
3  12  b(1)  c
 (1)

2
 5  5  b(5)  c (2)
 (3)
From (1), we have b  c  2
From (2), we have 5b  c  30  (4)
4b  32
b  8
(4) – (3),
By substituting b = 8 into (3), we have
8c  2
c  10
∵
∴
y = mx + k passes through the two points (1, 3) and (5, 5).
 (5)
3  m(1)  k

 5  m(5)  k  (6)
(6) – (5), 4m  8
m  2
By substituting m = 2 into (5), we have
2k  3
k 5
(c) By substituting b = 8 and c = 10 into y = x2 + bx + c, we have
(a)
y  x 2  8x  10
 y  x 2  8 x  10  (7)
Consider 
 (8)
3x  2 y  4
By substituting (7) into (8), we have
3 x  2( x 2  8 x  10)  4
3 x  2 x 2  16 x  20  4
2 x 2  13x  16  0
∴
The required quadratic equation is 2x2 – 13x + 16 = 0.
10. Find the minimum value of k so that the following simultaneous equations have real solutions.
 y  3x  8

2
 y  x  7x  k
##
......(1)
 y  3x  8

2
 y  x  7 x  k ......(2)
By substituting (1) into (2), we have
3x  8  x 2  7 x  k
x 2  4 x  (8  k )  0
......(3)
6
∵
∴
∴
The simultaneous equations have real solutions.
(3) has real roots.
0
∴
4 2  4(1)(8  k )  0
16  32  4k  0
4k  16
k4
The minimum value of k is 4.
11. It is given that the following simultaneous equations have real solutions.
 y  kx2  6

7 x  y  8  0
(a) Find the range of possible values of k.
(b) For the maximum value of k, solve the simultaneous equations.
##
 y  kx2  6
......(1)
(a) 
7 x  y  8  0 ......(2)
From (2), we have y = 7x  8 ......(3)
By substituting (3) into (1), we have
7 x  8  kx2  6
∵
∴
......(4)
kx2  7 x  14  0
The simultaneous equations have real solutions.
(4) has real roots.
0
∴
(7)  4(k )(14)  0
49  56k  0
56k  49
7
k
8
2
(b) From (a), the maximum value of k is
7
.
8
7 2

7
y  x  6
When k = , the simultaneous equations are 
.
8
8
7 x  y  8  0
By substituting k =
7
into (4), we have
8
7 2
x  7 x  14  0
8
x 2  8 x  16  0
( x  4) 2  0
x4
By substituting x = 4 into (3), we have
y = 7(4)  8 = 20
∴
When k =
7
, the solution of the simultaneous equations is (4, 20).
8
7
12. A wire 40 cm long is bent to form two squares. If the sum of their areas is 52 cm 2, find the lengths of the
sides of the two squares.
##
Let x cm and y cm be the lengths of the sides of the two squares respectively.
∴
4 x  4 y  40 ......(1)
 2
2
 x  y  52 ......(2)
From (1), we have y = 10  x ......(3)
By substituting (3) into (2), we have
x 2  (10  x) 2  52
2 x 2  20 x  48  0
x 2  10 x  24  0
( x  4)( x  6)  0
x  4 or x  6
By substituting x = 4 into (3), we have
y = 10  4 = 6
By substituting x = 6 into (3), we have
∴
y = 10  6 = 4
The lengths of the sides of the two squares are 4 cm and 6 cm respectively.
13. The perimeter of a rectangle is 42 cm. The product of the lengths of its two diagonals is 225 cm 2. Find the
length and the width of the rectangle.
##
Let x cm and y cm be the length and the width of the rectangle respectively.
∴
......(1)

2 x  2 y  42
2
 2
2
 225 ......(2)

 x y


From (1), we have y = 21  x
......(3)
From (2), we have x2 + y2 = 225 ......(4)
By substituting (3) into (4), we have
x 2  (21  x) 2  225
2 x 2  42 x  441  225
2 x 2  42 x  216  0
x 2  21x  108  0
( x  9)( x  12)  0
x  9 or x  12
By substituting x = 9 into (3), we have
y = 21  9 = 12
By substituting x = 12 into (3), we have
∴
y = 21  12 = 9
The length and the width of the rectangle are 9 cm, 12 cm or 12 cm, 9 cm respectively.
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