inequalities(2)

∵the equation has real roots
∴△≥0
[-(k+2)]^2-4k^2≥0
k^2+4k+4-4k^2≥0
3k^2-4k-4≤0
(3k+2)(k-2)≤0
-2/3≤k≤2

2011-11-29 22:23:01 補充：
also ∵it is the quadratic equation
∴k≠0
Hence the range of value of k is:-2/3≤k≤0 or o≤k≤2.

Answer 2/3<=k<0 or 0<=2 is wrong , what's 0 <=2 ?

2011-11-29 22:11:12 補充：
I think the answer should be -2/3 ≤ k < 0 or 0 < k ≤ 2

有無人幫計個正確答案出黎==

2011-11-29 22:21:05 補充：
yes the ans is -2/3 ≤ k < 0 or 0 < k ≤ 2
唔知點解打唔到出黎

2011-11-29 22:21:52 補充：
但係點計出黎呢??????????

Becos the question states that it is a "quadratic equation", so the coeff. of x^2 cannot be zero.

If k = 0, then it becomes -2x + 1 = 0 which is a LINEAR BUT NOT QUADRATIC equation.

For any quadratic equations Ax^2 + Bx + C = 0
x = [-B + sqrt (B^2 - 4AC)] / 2A and [-B - sqrt (B^2 - 4AC)] / 2A
if B^2 - 4AC < 0, there will be no real solution for x
if B^2 - 4AC = 0, there will be only one real solution for x i.e. -B/2A
if B^2 - 4AC > 0, there will be two real solutions for x

(k^2)(x^2) - (k+2)x + 1 = 0 has real roots
so B^2 - 4AC >= 0
(k+2)^2 - 4(k^2)(1) >= 0
k^2 + 4k + 4 - 4(k^2) >= 0
3(k^2) - 4k - 4 >= 0
(3k + 2)(k - 2) >= 0
case i
(3k + 2) >=0 and (k - 2) >= 0
k >= -2/3 and k >= 2
k >= 2
case ii
(3k + 2) <=0 and (k - 2) <= 0
k <= -2/3 and k <= 2
k <= -2/3

k <= -2/3 or k >= 2
(if no real root for x, the answer is exactly your one)

2011-11-29 10:46:28 補充：
sorry, my mistake

k^2 + 4k + 4 - 4(k^2) >= 0
3(k^2) - 4k - 4 <= 0
(3k + 2)(k - 2) <= 0
case i
(3k + 2) <=0 and (k - 2) >= 0
k <= -2/3 and k >= 2
no real solution for k
case ii
(3k + 2) >=0 and (k - 2) <= 0
k >= -2/3 and k <= 2
-2/3 <= k <= 2

但係就算k=0都計到個x,
x=0.5,
點解k唔可以=0?

2011-11-29 20:37:29 補充：
RE飛天魏國大將軍張遼 :
thx

k cannot be zero since it will make the coefficient of x^2 zero which implies that it is not a quadratic function.