EIV 4th ed Guide 3, Chapter 12 - pg 190

[ooisuankim]

In question 9, I agree that the answer is C. However I don't understand how Statement 2 is stated as "x^2 > x, so x<-1 OR x>1"

The answer I got was: x > 1 (when x > 0) and x < 1 (when x < 0).

Please clarify how is it possible to get x < -1.

Thanks.

[esledge]

There is a misprint (which we'll corrrect!), though your solution for (2) is not correct, either.

Algebraically, you can do the following with statement (2):
x^2 > x
(x^2) - x > 0
(x)(x - 1) > 0

A product is positive when you have pos*pos or neg*neg, so
(x)(x - 1) is pos*pos when x > 1.
(x)(x - 1) is neg*neg when x < 0.
The solution is therefore x > 1 or x < 0.

You could also test values, either to determine the ranges of values that "work" or to verify the solutions above:
For x = -10, x^2 = 100. 100>-10, so x^2 > x. OK
For x = -1/2, x^2 = 1/4. 1/4 > -1/2, so x^2 > x. OK
For x = neg, x^2 = pos. pos > neg, so x^2 > x. OK
For x = 1/2, x^2 = 1/4. 1/4 < 1/2, so x^2 is not > x. NO
For x = 1, x^2 = 1. 1 = 1, so x^2 is not > x. NO
For x = 2, x^2 = 4. 4 > 2, so x^2 > x. OK

The answer I got was: x > 1 (when x > 0) and x < 1 (when x < 0).

Please clarify how is it possible to get x < -1.

Abbout your logic, I'll point out that when you have x < 1 as your solution for the x < 0 case, there's some conflict: 0<x<1 are not negative solutions!

[ooisuankim]

Thanks, Ms Sledge. Does it mean that the answer for question 9 is still C?

i.e. if I combine Statement 1: x > 0 and statement 2: x > 1 or x < 0, do I get:

x > 0
and
x > 1

which means that the answer to the question x^3 > x^2 is a yes, as x is positive.


Thank you.

[Ben Ku]

You are correct. The answer is still C.