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(3 − x) (3+ x) ̧
(x + 6) (x + 1)
16-
$ 80
32k
x(x− 3)(x+ 1)
頁二第張六第日八廿月十年丑発展夏
1974中學會考試題預習專欄
「數學科(課程乙)(三) 女長波
MATHEMATICS (3)
Solution to exercise 2.
7.
a) 2(5x-2)(3-5x)
15(7 5x)
← 2(-25x2 + 25x
-50x2
5)
15(-5x + 2x 7)
50x
WAH KIU YAT PO
x(x + 6)
-x+3
provided:
x
3 x
3x
5x
(p - 1)(q - 1) 2pq (p+ q)
pq - (p + q) + 1
2()()
The solution set of k is
{x
Example
• k < 2}
If p, q are the roots
of 2x -x+1 = 0; find a "quadratic equation such that.
the roots of this equation are
and
Solu
qware the
p
and roots of
-75x + 30x + 105
20x 117 -0
(5x
13)(5x - 9)
-2
The solution set of x is
(-23 1.
- 4(2)(−3)
34/33
The solution set of
2. Let x be the tens digit of the
number, then the units digit of the number is (x - 3)
10x + (x - 30
11x
台
2x2
8)(2x
(x -.3)2 -
6x + 9
17% + 8 =
(2x+1)(2x+1
7x+
X
(2x+ 1)(x-2)
5x
(2x+ 1)(x+ 3)
1) (2x+1)= (x+3)(x+3)+5x(x-2
*4x2+4x+1mx2+6x+9+5x210x
(x+3)(x-2)(2x+1)
8x .:::12x-8 (x+3)(x-2)(2x+ 1)
4(2x+1)(x-2) (x+3)(x−2)(2x+ 1)
provided x
Sun and Product of the roots of the quadratic equation.
If pandq are the roots of the quadratic equation:
ax?
+ bx + c = 0, we have:
Sum of rootst P
2. Product of roots:
Example 1. If p and q are
roots of 2 - 4x + 5 = 0,
find, without solving the "equation, the values of
(rejected)
Solution: Since
and
8 or
x-3= 8 3 =
The number is 854
2x2 - 3x
7x+6
(3-x)(3+x)(x+6)(x+1)
(x+6)
x(x-2x-3)
款金
(p+q) (p? + q? - pq.) ()(-13- ) ()(-22)
427
Example 2. In each quadratic
equation, find the solution set of the constant k subject to the. give condition
3x2
2
3kx + k
80:
has the sum of the roots equal: to the product of its roots.
4x2 + 2k = 8x +
has two unequal real roots.
Solution.
a) Sum of roots
product of roots
The solution set
k is
2k 8x +
> 4x2 - 8x + 2k
Since the equation has two un-
equal real roots,
Discriminant
(-8)2- 4(4)(2k
(1)2 - 2(1)
The required equation is
)x + 1
xercise 3.
:2pq
ts. If_p, q are the roots of
+510x, find without solving the equation, the value of:
a) (p
9) 2
b)(p - 2q)(2p c) [p = q|
2. In each quadratic equation, i
find the value(s) of k subject to the given condition. Hence, find the Solution set of each quadratic equation.
a) 2x2 - (k+1)x = 1
has the sum of its roots:
"equal to twice the producet
of its roots.
b) 6x2 + 5x + k
has equal roots.
Find a quadratic equation such that the roots are reciprocals
1)70
of the roots of
3x
2x + 5
+
({)2 - 2(})
64- 32k + 16>0
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