Hi Chris,

x^2 is greater than x if x is bigger than 1 or if x is any negative value (since x^2 would be positive).

The rephrase should be "Is x >1 or negative?"


(2) is not sufficient because knowing that x > -1 doesn't tell us whether x > 1 or negative. X could be 0.5 (x^2 would be smaller than x) but x could also be 2 (x^2 is bigger than x)

Is X^2 greater than X?
(x^2 -x) > 0----------------------(A)
1)X^2 is greater than 1
2) X is greater than -1

to prove: x^2 - x > 0

1) gives us x^2 -1 > 0
=> x > 1 or x < -1
for x > 1
x^ 2 - x > will always be true ,take any value fraction or integer where value > 1

example x = 2
4 -2 > 0
example 3/2
9/4 - 3/2 => 9-6/4 = 3/4 > 0

for x < -1
example -2
4 -(-2) = -6

example -5/2

25/4 -(-5/2) > 0

2) gives us x > -1

which means x can be 0,1/2, 1 2, 3/2

lets see for what values the equation (A) works
example 2 : 4-2 > 0

example 1/2
1/4 -1/2 < 0

Hence (A)

s2 is not suff
x>-1=-0.5;1;2(for each of these x^2 and x have different relationship)

Patrick_GMATFix wrote:Hi Chris,

x^2 is greater than x if x is bigger than 1 or if x is any negative value (since x^2 would be positive).

The rephrase should be "Is x >1 or negative?"


(2) is not sufficient because knowing that x > -1 doesn't tell us whether x > 1 or negative. X could be 0.5 (x^2 would be smaller than x) but x could also be 2 (x^2 is bigger than x)

Hi Patrick,

I get the why now statm 2 is not sufficient, but i don't understand your rephrase; from this X(X-1)>0 I get 2 solution X>0 and X>1, so both showing that X is "requested" to be positive. Why should be asking that X is negative too? What am I missing here?

Thx

ironsferri wrote:

Patrick_GMATFix wrote:Hi Chris,

x^2 is greater than x if x is bigger than 1 or if x is any negative value (since x^2 would be positive).

The rephrase should be "Is x >1 or negative?"


(2) is not sufficient because knowing that x > -1 doesn't tell us whether x > 1 or negative. X could be 0.5 (x^2 would be smaller than x) but x could also be 2 (x^2 is bigger than x)

Hi Patrick,

I get the why now statm 2 is not sufficient, but i don't understand your rephrase; from this X(X-1)>0 I get 2 solution X>0 and X>1, so both showing that X is "requested" to be positive. Why should be asking that X is negative too? What am I missing here?

Thx

x(x-1)>0
the above is true not only for bth +ve , but also for bth -ve

thephoenix wrote:

ironsferri wrote:

Patrick_GMATFix wrote:Hi Chris,

x^2 is greater than x if x is bigger than 1 or if x is any negative value (since x^2 would be positive).

The rephrase should be "Is x >1 or negative?"


(2) is not sufficient because knowing that x > -1 doesn't tell us whether x > 1 or negative. X could be 0.5 (x^2 would be smaller than x) but x could also be 2 (x^2 is bigger than x)

Hi Patrick,

I get the why now statm 2 is not sufficient, but i don't understand your rephrase; from this X(X-1)>0 I get 2 solution X>0 and X>1, so both showing that X is "requested" to be positive. Why should be asking that X is negative too? What am I missing here?

Thx

x(x-1)>0
the above is true not only for bth +ve , but also for bth -ve

Thank you Thephoenix. I know x^2 can hide the sign, but when I see this form (x(x-1)>0) I don't see it. I think it's more a basic math issue at this point, but still want to understand it. why in x(x-1)=0 I have x=0,1 and in x(x-1)>0 is x>0 and x<0 ? Would you mind elaborate on how to reach that result?

Thank you

If x(x-1) > 0, what can we infer about the value of x ?


Note that x(x-1) is a quadratic expression that can be written in the form ax^2 + bx + c, and in this case a= 1
Thus means that y=x(x-1) is a u-shaped (concave) parabola. Thus x(x-1) > 0 if x < 0 or x > 1


Alternatively,

x(x - 1) > 0

Let's simplify by dividing both sides by x (assuming x is not 0)


If x > 0, we get x - 1 > 0, i.e. x > 1

If x < 0, we get x - 1 < 0, i.e. x < 1

Thus x > 1 or x < 0

ironsferri wrote: Thank you Thephoenix. I know x^2 can hide the sign, but when I see this form (x(x-1)>0) I don't see it. I think it's more a basic math issue at this point, but still want to understand it. why in x(x-1)=0 I have x=0,1 and in x(x-1)>0 is x>0 and x<0 ? Would you mind elaborate on how to reach that result?

Thank you

Hello ironsferri,

You're right to rewrite x^2 > x as (x)(x-1)>0. For two values to have a positive product, they must have the same sign, so either:

1) they're both positive: x >0 and
x-1 >0 --> x>1

In this case, x>0 and x>1 can just be written as x>1 (because this is where both inequalities are true)

2) they're both negative: x<0 and
x-1>0 --> x>1
Because in this case x is negative, we can ignore the 2nd inequality (x cannot be bigger than 1) and write this case as x<0

Thus when you include both cases in a single statement, you get: x>1 or x<0. That's the proof for my rephrase above.

The way I rephrased however was to simply think about which values of x would make "is x^2>x?" true.

• I know that for any negative x, x^2 will always be greater since it will be positive.

• I also know that for "regular" numbers (2, 3, 4, 5...), x^2 is greater.

• However because positive fractions get smaller when you square them, "is x^2>x?" would not be true for positive fractions.

• Finally I know that for x=1 or 0, x^2 would not be greater than x

In short the only way to make the answer be a NO is to make x a positive fraction. So either of the rephrases below would work: "Is x<0 or x>1?" OR "Is 0<= x <= 1?"

Good luck,
-Patrick