urgent add maths~~~~~inequalities

(a) Discriminant
= [2(m - 4)]^2 - 4(1 - 2m)(4 - m)
= 4(m - 4)^2 - 4(1 - 2m)(4 - m)
= 4(m^2 - 8m + 16) - 4(4 - 9m + 2m^2)
= 4m^2 - 32m + 64 - 16 + 36m - 8m^2
= -4m^2 + 4m + 48
= -4(m^2 - m - 12)
= -4(m - 4)(m + 3)
(b) (i) f(x) > 0 for all real values of x. Thus the graph y = f(x) opens upwards and does not cut the x-axis.
i.e. (1 - 2m) > 0 and -4(m - 4)(m + 3) < 0
1 > 2m and (m - 4)(m + 3) > 0
m < 1 /2 and [m > 4 or m < -3]
m < -3
(ii) f(x) < 0 for all real values of x. Thus the graph y = f(x) opens downwards and does not cut the x-axis.
i.e. (1 - 2m) < 0 and -4(m - 4)(m + 3) < 0
1 < 2m and (m - 4)(m + 3) > 0
m > 1 /2 and [m > 4 or m < -3]
m > 4

Discriminant = [2*(m - 4)]^2 - 4*(1 - 2m)*(4 - m)
= 4*(m^2 - 8m + 16) - 4* (4 - 9m + 2m^2)
= 4*m^2 - 32m + 64 - 16 + 36m - 8*m^2
= - 4*m^2 + 4m + 48
= - 4*(m^2 - m - 12)
= - 4*(m-4)*(m+3)

(1) f (x) is greater than 0 for all real values of x,
f (x) &gt; 0, (1-2m) &gt; 0
m &lt; 1/2 and
it is no real root,
So discriminant &lt; 0,
- 4*(m-4)*(m+3) &lt; 0
(m-4)*(m+3) &gt; 0
m &lt; -3 or m &gt; 4
Therefore, m &lt; -3

(2) f (x) is less than 0 for all real values of x,
f (x) &lt; 0, (1-2m) &lt; 0
m &gt; 1/2 and it is no real root,
So discriminant &lt; 0,
- 4*(m-4)*(m+3) &lt; 0
(m-4)*(m+3) &gt; 0
m &lt; -3 or m &gt; 4
Therefore, m &gt; 4