Is |x|^2 - 5|x| + 6 0? (1) |x| 2 (2) |x|

Question

Is  |x|^2-5|x|+6>0?

(1)  |x|>2

(2)  |x|<3

GMATinsight · Answer

Let, |x| = a |x|^2-5|x|+6>0 becomes a^2-5a+6>0 i.e. Question is Is (a-2)(a-3) > 0? Which is possible when both (a-2) and (a-3) are greater than 0 i.e. Question : Is a > 3 OR a<2? cause a=IxI can NOT be negative Statement 1: a > 2 NOT SUFFICIENT Statement 2: a < 3 NOT SUFFICIENT Combining the two statements 2 < a < 3 SUFFICIENT Answer: Option C

Archit3110 · Answer

#1
at x= 3 , we get  |x|^2-5|x|+6>0?  = 0
and x=4  |x|^2-5|x|+6>0? = 2 ;insufficient
#2
x=2,1  |x|^2-5|x|+6>0? =0 
insufficient

from 1&2
2>x<3
sufficient
IMO C

stne · Answer

In inequality it is better to verify the answer by plugging in the original equation.

1) if x=3 then  |x|^2-5|x|+6 =0  ans is NO.
 if x=4 then  |x|^2-5|x|+6 =2  ans is Yes 
Hence Insufficient

2) If x=2  |x|^2-5|x|+6  =0 ans is NO.
 if  x=0  then  |x|^2-5|x|+6  =6 ans is Yes 
Hence B is also Insufficient

1+2

-3< x <-2 or 2<x<3 
take a value x= \frac{5}{2}   \rightarrow  2.5 putting in eqn.
 \frac{25}{4}-\frac{25}{2}+6<0 
So a definite NO.
IMO C

TheUltimateWinner · Answer

GMATinsight 
Here the explanation of statement 2 says: B is Insufficient.

TheUltimateWinner · Answer

Archit3110 
Take a look here, please.

Kapilg1 · Answer

what if we take a value of -5/2 
then this inequality is satisfied and for 5/2 its not. hence answer should be E

Bunuel · Answer

You can check correct answer under the spoiler in the original post. For this question it's C, not E.

When we combine the statements we get: 2 < |x| < 3. For all such value of |x|, |x|^2 - 5|x| + 6 is negative. So, when we combine the statements we can give a definite NO answer to the question. So, the answer is C.

Hope it helps.

Fdambro294 · Answer

(1st) since Squaring a Number always results in a (+)Positive Output, the Modulus under the Square is redundant.

Q stem becomes:

Is: (x)^2 - 5*[x] + 6 > 0?

(2nd) what are the Critical Values that makes the expression 0

X = (-)3 ; (-)2 ; +2 ; +3

And we are looking for which Ranges:

Is: (x)^2 - 5[x] + 6 > 0 ?

Is: (x)^2 + 6 > 5[x] ?

(3rd) Plotting the Critical Values and finding the Ranges that would deliver a YES or a NO answer

— (-)3—— (-)2 —- +2 ——— +3 —

If x = anyone of the Critical Values above, the Inequality will be EQUAL and the answer will be NO

x > + 3:
If x = 4———> (4)^2 + 6 > 5[4]
YES

2 < x < 3:
If x = 2.5 ——> 6.25 + 6 < 12.5
NO

(-)2 < x < 2:
If x = 0 ——> 0 + 6 > 5 * [0]
YES

(-)3 < x < (-)2
If x = (-)2.5 ——> 6.25 + 6 < 12.5
NO

x < (-)3:
If x = (-)4 —> (-4)^2 + 6 > 5 * [-4]
YES

Summary:

Case 1:

If: (-)3 </= x </= (-)2

or

+2 </ = x </= +3

Answer is NO

Case 2: for all other real values of X, answer is YES

Stmt. 1
[x] > 2 ——> x > +2 OR x < (-)2

Yes and No possible. NOT Suff

Stmt.2
[x] < 3 ——> (-)3 < x < +3

Yes and No possible. NOT Suff

Together

Combining the Inequalities and only accounting for Ranges that are COMMON to BOTH Inequalities (in other words, where the Inequalities OVERLAP on the Number Line)

On (-)Negative Side of the No Line:

(S1) x < (-)2
(S2) x > (-)3

(-)3 < x < (-)2

On (+)Positive Side of the No. Line:

(S1) x > +2
(S2) x < + 3

+2 < x < +3

Therefore, the statements together tells us that the only possible Ranges of X Values are:

(-)3 < x < (-)2 OR +2 < x < +3

As shown above in the Question Prompt Analysis, any x value in these Ranges will always return a NO Answer to the Question.

C- together statements are sufficient

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asadL · Answer

Won't we get a>3 and a>2 instead of a<2, where am I mistaken?

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CEdward · Answer

Tough one. I am not sure if the way I did it is the fastest. But it's thorough?

Is |x|^2−5|x|+6>0?

This can be rewritten as (|x| - 3)(|x| - 2) > 0? We have two options:
a) Both terms are positive in which case x > 4 or x < -3
b) Both terms are negative in which case x ≥ -2 or x < 2

(1) |x|>2 --> x > 2 or x < -2
Insufficient considering the options above

(2) |x|<3 --> x < 3 or x > -3
Insufficient considering the options above.

Combo pack:
2 < x < 3
-3 < x < -2

So x can be -2.5 or 2.5 for example. Both suggest that (|x| - 3)(|x| - 2) IS NOT > 0.

Sufficient.

Oppenheimer1945 · Answer

Take |x|=t,
Is (t-2)(t-3)>0?

We know the exp is -ve when t lies b/w 2&3, and +ve otherwise as per wavy method or number line.

S1) NS
S2) NS
S1+S2) exp is negative

C)

Is |x|^2 - 5|x| + 6 > 0? (1) |x| > 2 (2) |x| < 3

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