有5題maths a0a 好易 Q,2,4,5 XXX

回答 (2)

2007-12-06 8:39 am
✔ 最佳答案
[Set 1]
1.
(b)
(2x+3)^2 - 25 = 0
(2x+3)^2 = 25
2x+3 = ±5
2x = -3±5
x = 1 or x = -4

[Set 2]
1.
(a)
Discriminant
= (7)^2 - 4(5)(-k)
= 49 + 20k
(b)
The equation has two equal real roots, discriminant = 0.
49 + 20k = 0
20k = -49
k = -49/20

[Set 3]
1.
Discriminant
= (3)^2 - 4(4)(-k)
= 9 + 16k
If the equation has no real roots, discriminant < 0.
9 + 16k < 0
16k < -9
k < -9/16

2.
Let the two positive even numbers be x and x+2.
x(x+2) = 288
x^2 + 2x - 288 = 0
(x - 16)(x + 18) = 0
x = 16 or x = -18 (rejected)
The two numbers are 16 and (16+2), that is 18.

If there is a mistake, please inform me.
參考: My Maths knowledge
2007-12-05 9:38 pm
B
(2x+3)^2-25=0
(2x+8)(2x-2)=0
2x+8=0 OR 2x-2=0
x=-4 OR x=1


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