[2]一些不等式

5.a) 判別式 = (2k)^2 - 4(1)(6+k) > 0
4k^2 - 4k - 24 > 0
k^2 - k - 6 > 0
(k - 3)(k + 2) > 0
k > 3 或者 k < -2
b)想知你要一個正數根 or 一個負數根、一個重覆的正數根 或 兩個都係正數根呢?????
(i) 如果係兩個正數根，
2k 一定要負數, 因 (x - a)(x - b) 先會出到 正數根，而佢coefficient of x 就會係負數，
所以， k < 0 。
同時 6 + k > 0，因constant term 要係正數先會令 (-a)(-b)的出現，所以，k > -6
最後得出 -6 < k < 0.
(ii) 如果係一個重覆的正數根，判別式一定要等於 0。
k^2 - k - 6 = 0
(k - 3)(k + 2) = 0
k = 3 或 k = -2
但由於 k < 0 ，所以 k = -2
(iii) 如果係一個正數根 or 一個負數根，判別式就如5a) 一樣的k可取的實數值範圍。
但 6 + k < 0，所以 k < -6
最後，k < -6 。

6. r = (x-1) / (x^2 + x + 1)
r(x^2 + x +1) = x-1
rx^2 + (r-1)x + (r+1) = 0
判別式 >= 0
(r-1)^2 - 4(r)(r+1) >= 0
r^2 - 2r + 1 - 4r^2 - 4r >= 0
3r^2 + 6r - 1 <= 0
所以， -1 - 2/sqr(3) <= r <= -1 + 2/sqr(3)

7.a) y = -x / (x^2 - 2x + 4)
y(x^2 - 2x + 4) = -x
yx^2 - 2yx + 4y +x = 0
yx^2 + (1-2y)x + 4y = 0
判別式 >= 0
(1-2y)^2 - 4y(4y) >= 0
4y^2 - 4y + 1 - 16y^2 >= 0
12y^2 + 4y - 1 <= 0
所以，(-2-sqr(15))/12 <= y <= (-2+sqr(15))/12
b) y 的最大值就是(-2+sqr(15))/12，那麼就由此去找x。

5) x^2 + 2kx + (6+k) =0
a) To have 實數根,
Determinant=(2k)^2-4(6+k)is greater than or equals to 0
k^2-k-6 is greater than or equals to 0
(k+2)(k-3) is greater than or equals to 0
k is smaller than or equals to -2 or k is greater than or equals to 3

b)To have 正數根,
Sum of root and product of root must both are positive, i.e. greater than 0

So, sum of root = -2k is greater than 0
k is smaller than 0
product of root = 6+k is greater than 0
k is greater than -6
but 正數根 must also be 實數根
So, we need to consider part a's result
Combining all the conditions, -6 is smaller than k is smaller than or equals to -2

6) r = (x-1)/(x^2+x+1)
(x^2+x+1)r = x-1
rx^2+(r-1)x+r+1 = 0

For x is 任意實數值
Determinant = (r-1)^2-4r(r+1) is greater or equals to 0
r^2-2r+1-4r^2-4r is greater or equals to 0
3r^2+6r-1 is smaller or equals to 0
[-6-{36+12}]/6 is smaller than or equals to r smaller than or equals to [-6-+36+12}]/6 , where {} is square root.

[-6-4{3}]/6 is smaller than or equals to r smaller than or equals to [-6+4{3}]/6
-2{3}/3 - 1 is smaller than or equals to r smaller than or equals to 2{3}/3 - 1

7) y = -x/x^2-2x+4，且x為實數
a) y(x^2-2x+4) = -x
yx^2+(1-2y)x+4y = 0
Determinant = (1-2y)^2 - 4y(4y) is greater than or equals to 0 for x為實數
1-4y+4y^2-16y^2 is greater than or equals to 0
12y^2+4y-1 is smaller than or equals to 0
(2y+1)(6y-1) is smaller than or equals to 0
-1/2 is smaller than or equals to y is smaller than or equals to 1/6
b) y之值is最大, that means take y=1/6
1/6 x^2 + (1-1/3) x + 4 (1/6) = 0
x^2+4x+4=0
(x+2)^2=0
x=-2