一元二次方程

3x^2+3x-k=0 有兩個相等的實根,求k的值,,,(書答案-3/4)
判別式 3^2 - 4(3)(-k) = 0
9 + 12k = 0
k = -3/4
(k-1)x^2+4x-7 =0 有兩個相等的實根,求k的值
判別式 4^2 - 4(k-1)(-7) = 0
16 +28k - 28 = 0
28k = 12
k = 3/7
2x^2 - 4k + k = 0有兩個不等的實根,求k的所有可取值
判別式 4^2 - 4(2)(k) > 0
16 - 8k > 0
16 > 8k
2 > k
5x^2 - 4(x+k) - 12 = 0有兩個不等的實根,求k的取值範圍.
5x^2 - 4x - 4k - 12 = 0
5x^2 - 4x - (4k+12) = 0
判別式16 - 4(5)[-(4k+12)] > 0
16 + 20(4k+12) > 0
16 + 80k + 240 > 0
80k > -256
k > -3.2

2009-11-13 23:31:42 補充：
y= -5x^2 -x +k 判別式大於 0
1 + 20k > 0
20k > -1
k > -1/20
你做的錯了:1 > -20k => -1/20 < k 當左右乘除負數時,不等號要改變方向.

y= -5^2 -x +k 判別式大於 0

1-(-20)k>0

1 > -20k

1 / -20 >k

但答案係 k>1/-20...有咩方法令條式靚d?

要教你怎麼釣魚,而不是給你魚,不然就跟給你答案,但是你不會自己思考答案怎麼來的?一樣,若是這樣,會害死你.而不是幫助你.

since it has 2 equal real roots
so &Delta;=0
&Delta;=b^2-4ac
b^2-4ac=0
3^2-4(3)(-k)=0
9+12k=0
k=-3/4

since it has 2 equal real roots
so &Delta;=0
b^2-4ac=0
4^2-4(k-1)(-7)=0
16-(4k-4)(-7)=0
16-(-28k+28)=0
16+28k-28=0
28k-12=0
k=3/7

since it has 2 distinct real roots
so &Delta;&gt;0
b^2-4ac&gt;0
(-4)^2-4(2)(k)&gt;0
16-8k&gt;0
8k&lt;16
k&lt;2

since it has 2 distinct real roots
so &Delta;&gt;0
5x^2-4x-4k-12=0
5x^2-4x-4(k+3)=0
(-4)^2-4(5)﹝-4(k+3)﹞&gt;0
16-20(-4k-12)&gt;0
16+80k+240&gt;0
80k&gt;-256
k&gt;-32