If x^2 is a positive integer, is x a positive integer?

If X^2 is a positive integer, is x a positive integer?

(1) If x = 2, then √x^2 = 2, so x = +ve integer.
If x = -2, then √x^2 = 2, So, x = -ve integer.
So, insufficient.

(2) √x^2 = x,
If x = 2, then √2^2 = 2, so x = +ve integer.
If x= -2, then √2^2 = -2, Therefore, x can not be -ve integer.
If x = 1.1, then √1.1^2 = 1.1 but x = -1.1, then √-1.1^2 = 1.1
Therefore, x can never be -ve integer. So, x can always be +ve integer.

Imo. B

x2>=1
therefore either x>=1 or x<=-1.
q: is x>=1?
a)√x2 is an integer. therefore √x2>=1
squaring x2>=1, which is same as the information provided. inconclusive
b) √x2=x
squaring: x2=x2. inconclusive.
IMO E

From stmnt (1),
√x^2 = 2, or,x^2 = 4, or, x = ±2, So A and D are out.

From stmnt (2), √x^2 = x, or,x^2 = x^2

From the two stmnts, we do not know the sign of x.(E)

C is the answer as
√x = mod x ....so X can be positive and negative both


But if √x = x means x can be positive only as value in root can't he negative...

But √x can also be √2 in that case it's not an integer .... therefore we require statement 1 also

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If x^2 is a positive integer ,is x a positive integer ?

Constraint: x^2= +ve Integer
so x can’t be 0
Question: is x>0 Yes? Or
x<= 0 No?

(1) Sqrt.(x^2) = Integer
|x| = Integer, basically says the distance between 0 and x is an integer.
When x=-5 ,the distance is 5 units from zero
When x=5 ,the distance is also 5 units
-5 <———0——-> 5
x= +ve /-ve. (Not sufficient)

(2) Sqrt.(x^2)=x
|x| =x ,since the RHS is +ve
x >= 0 but x can’t be zero
So x>0 , x=+ve
(Sufficient )

Hit that B

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Given x^2 is a positive integer
--> x can be positive/negative integer or positive/negative irrational number
--> A possible set of values of x = ±\(2\), ±\(\sqrt{2}\)

(1) \(\sqrt{x^2}\) is an integer.
--> Possible value of x = ±2 --> Insufficient

(2) \(\sqrt{x^2} = x\)
--> 'x' has to be positive
--> Possible value of x = \(2\) or \(\sqrt{2}\) --> Insufficient

Combining (1) & (2),
--> Possible value of x = 2 ONLY (Or any positive integer only) --> Sufficient

Option C

Given that x^2 is a positive integer, we are to determine if x is a positive integer.

Statement 1: √x^2 is an integer
Not sufficient. For example, if x^2=4, then x can be 2 or -2, since 2^2=4 and (-2)^2=4. Since both -2 and 2 are both integers, with -2 being a negative integer and 2 being a positive integer, we cannot determine whether x is a positive integer or not based on statement 1.

Statement 2: √x^2 = x
Not sufficient. x can be positive or negative. x is just a variable. If x^2=4, then x can be -2 or 2. Since x can either be negative or be positive, we cannot determine whether x is positive or not.

1+2
Both statements even when combined is not sufficient to determine if x is positive or negative.

The answer is E.

If \(x^2\) is a positive integer, is x a positive integer?
x > 0 ?

(1) \(√x^2\) is an integer.
if \(x^2\) = 4,
\(√x^2\) = +2 YES
or
\(√x^2\) = -2 NO

INSUFFICIENT.

(2) \(√x^2 = x\)
Squaring
\(x^2 = x^2\)

Nothing further given
INSUFFICIENT.

Together 1 and 2.
No new information.

IMO Answer E.

\(x^2=positive.integer…|x|>0…x=[\sqrt{2},3,-4…]\)

(1) √x^2 is an integer insufic

\(\sqrt{x^2}=|x|=integer…x=[-3,5…]\)

(2) √x^2=x insufic

\(\sqrt{x^2}=x…\sqrt{x^2}=|x|=x…|x|=x…|x|>0…x>0…x=[\sqrt{2},5…]\)

(1&2) sufic

\(x=integer>0\)

Ans (C)

If \(x^{2}\) is a positive integer, is x a positive integer?
(—> x cannot be equal to zero.)

(Statement1): \(√x^{2}\) is an integer.
—> x could be either positive or negative integer
If x=—3, then \(√(—3)^{2}\) =integer (No)
If x= 3, then \(√3^{2}\) =integer (Yes)
Insufficient

(Statement2): \(√x^{2} = x\) (—> x >0)
If x= 3, then \(√3^{2}\) =3 (Yes)
If x= √3, then \(√(√3)^{2}\) =√3 (No)
Insufficient

Taken together 1&2,
\(√x^{2} = x\) and x is an integer.
Only if x is an positive integer, it satisfies the both statements
Sufficient

The answer is C

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If x^2 is a positive integer, is x a positive integer?

(1) √x^2 is an integer.
√x^2 = |x|
x can be positive or negative integer.
Insufficient

(2) √x^2=x
As √x^2 = |x|
Therefore, |x| = x;
x is positive number, but may or may not be an integer.
Insufficient

(1)+(2), x is an integer which is positive
Sufficient

C is correct

Ans: C

x^2 is a positive integer-->> 5,9 etc
(1) √x2 is an integer...so x^2 cannot be 5
it can be 9..so x= +3 or -3. not sufficient

2)√x2=x, so x can be √5 or 3. not sufficient

combined-->> x can only be integer and positive

If \(x^2\) is positive, \(x\) is not equal to 0

(1) \(\sqrt{x^2}\) is an integer.

\(\sqrt{x^2}\)=\(|x|\)=integer

If \(|x|\) is an integer and \(x\) is not 0, \(|x|\) must be a positive integer

So \(x\) is a positive integer if \(|x|=x\)

\(x\) is a negative integer if \(|x|=-x\)

1 is not sufficient

(2) \(\sqrt{x^2} = x\)

Here, \(|x|=x\) so \(x\) must be positive but we don't know if it is an integer

2 is insufficient

(1)+(2)

\(|x|\) is an integer and \(|x|=x\) so \(x\) must be a positive integer

Sufficient

Answer is (C)

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Damn it! subtle stuff,was all about integers and non-Integer. Lolx why I love this 700-level Questions

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Exact emotions here as well .
Somewhere while solving had a hunch that i was going wrong.

Asked: If \(x^2\) is a positive integer, is \(x\) a positive integer?
Since |x| is a positive integer; x<>0

(1) \(\sqrt{x^2}\) is an integer.
|x| is an integer
x may or may not be positive
NOT SUFFICIENT

(2) \(\sqrt{x^2} = x\)
|x| = x
Since |x|>0; x>0
x is a positive number but may not be an integer
NOT SUFFICIENT

(1) \(\sqrt{x^2}\) is an integer.
|x| is an integer
(2) \(\sqrt{x^2} = x\)
|x| = x
Since |x|>0; x>0
x must be a positive integer
SUFFICIENT

IMO C

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