1973-11-22 — Page 22

報日僑華

四期車

日二廿月一十年三七九一层公年二十六國民華中

(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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