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

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

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

Archit3110
Take a look here, please.

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

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.

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

Posted from my mobile device

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.