Quodratic Equations數學題一問~

2009-04-08 5:45 pm
1)
If the product of two expressions (x + a) and (x - b) is zero, then what is true about the expressions ?

a)
a + b = 0

b)
x + a = 0 & x - b = 0

c)
x + a = 0 or x - b = 0

d)
a x b = 0

e)
None of the above


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回答 (4)

2009-04-08 9:50 pm
✔ 最佳答案
(x + a)(x - b) = 0



a) False


a + b = 0, ie. a = -b


(x + a)(x - b) = (x - b)(x - b) = (x - b)^2 >/= 0
(greater than or equal to 0)



b) False


x + a = 0 & x - b = 0, ie. x = -a & x = b


Let (x1 + a)(x2 - b) = 0,


If x1 = -a, x2 = b, (x1 + a)(x2 - b) = (-a + a)(b - b) = 0*0 = 0


If x1 = b, x2 = -a, (x1 + a)(x2 - b) = (b + a)(-a - b) = -(a + b)^2 </= 0
( smaller than or equal to 0)



c) True


x + a = 0 or x - b = 0, ie. x = -a or x = b


If x = -a, (x + a) = 0 => (x + a)(x - b) = 0


If x = b, (x - b) = 0 => (x + a)(x - b) = 0

=> (x + a)(x - b) = 0



d) False, as it can be greater than/ smaller than/ equal to 0


a x b = 0, ie. a = 0 or b = 0


Consider a = 0, (x + a)(x - b) = x(x - b)


When x > b, x >/= 0
x(x - b) >/= 0


You can prove others, eg.
When x > b, x < 0
x(x - b) < 0



e) False
None of the above

2009-04-08 13:55:39 補充:
d) when a = b = 0,

(x + a)(x - b) = x^2 >/= 0
2009-04-09 2:08 am
更正一下,應該係「Quadratic Equations」。(係『a』唔係『o』)
2009-04-08 6:22 pm
答案是 C.

解釋 :

當知道兩數字的積等於 0 時, 那麼便知道在這兩個數字中最少

其中一個一定是 0. 但是, 也有可能兩個數字都是 0.

不過, 在這裡, x 不可能同時等於 - a 和 b. 因此, 得出的結論是 :

x + a = 0 或 x - b = 0 (C), 而 x + a = 0 & x - b = 0 (B) 是不可能的.

另外, (A) 和 (B) 皆不是直接從問題的條件可得出的結論.

2009-04-08 10:31:40 補充:
更正 :

另外, (A) 和 (D) 皆不是直接從問題的條件可得出的結論.

2009-04-08 10:41:18 補充:
給 001 的朋友 :

雖然 "AorB係包含左A對或B對或A&B都對" 可能是對的,

但是在這裡, x 不可能同時等於 - a 和 b.

而且按你的意思, (B) 並不是一個錯的結論, 只是一個不完全的結論.
2009-04-08 6:09 pm
答案係C

因為AB=0, 咁一係A=0, 一係B=0, 一係A&B同時都=0 才能令AB=0
而AorB係包含左A對或B對或A&B都對
參考: ME


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