z(z-4) > 5

Lets take each option 1 by 1

a) z = 0 invalidates the above
b) z = 0 invalidates the above
c) z = 0 invalidates the above
d) z = 0 invalidates the above
e) All values of z for z < -1 keeps the above inequality

So,(E)

GmatKiss wrote:If z² - 4z > 5 then which of the following is always true

a)z > -5
b)z < 5
c)z > -1
d)z < 1
e)z < -1

Try to show that the answers DON'T have to be true.

Plug z=10 into z² - 4z > 5:
10² - 40 > 5
60 > 5.
This works.
Since it's possible that z=10, eliminate B, D and E.

Plug z= -10 into z² - 4z > 5:
(-10)² - 4(-10) > 5
100+40 > 5
140 > 5.
This works.
Since it's possible that z=-10, eliminate A and C.

None of the answers must be true.

Algebraically:
z² - 4z - 5 > 0
(z+1)(z-5) > 0.
Thus, z=-1 and z=5 are the critical points: the values where z² - 4z - 5 = 0.
When z is ANY other value, z² - 4z - 5 < 0 or z² - 4z - 5 > 0.
To determine the range of z, test values to the left and right of each critical point.

Case 1: z < -1.
Let z = -2.
(-2)² - 4(-2) - 5 > 0
7>0. This works.
z<-1 is part of the range.

Case 2: -1<z<5.
Let z=0.
(0)² - 4(0) - 5 > 0
-5>0.
Doesn't work.
-1<z<5 is not part of the range.

Case 3: z>5.
Let z=6.
(-6)² - 4(6) - 5 > 0.
7>0.
This works.
z>5 is part of the range.

Thus, z<-1 OR z>5 .
Since it's possible that z<-1 or that z>5, none of the answers MUST be true.

we can solve it directly plugin the values. It would be the simplest way.