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Is root{x} a prime number?

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Re: Is root{x} a prime number? [#permalink]

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New post 25 Dec 2015, 00:29
mohnish104 wrote:
Bunuel wrote:
LM wrote:
Is \(\sqrt{x}\)

(1) \(|3x-7|=2x+2\)

(2) \(x^2=9x\)


Is \(\sqrt{x}\) a prime number?

1) \(|3x-7|=2x+2\) --> we have one check point 7/3 (check point - the value of x for which an expression in absolute value equals to zero):

A. \(x\leq{\frac{7}{3}}\) --> \(3x-7\leq{0}\) hence \(|3x-7|=-(3x-7)\) --> \(-(3x-7)=2x+2\) --> \(x=1\) --> \(\sqrt{1}=1\neq{prime}\).
B. A. \(x>{\frac{7}{3}}\) --> \(3x-7>0\) hence \(|3x-7|=3x-7\) --> \(3x-7=2x+2\) --> \(x=9\) --> \(\sqrt{9}=3=prime\).
Two different answer. Not sufficient.

2) \(x^2=9x\) --> \(x(x-9)=0\) --> \(x=0\) or \(x=9\) --> \(\sqrt{0}=0\neq{prime}\) or \(\sqrt{9}=3=prime\). Not sufficient.

(1)+(2) Intersection of values from (1) and (2) is \(x=9\) --> \(\sqrt{9}=3=prime\). Sufficient.

Answer: C.





Bunuel, I have a question. Usually while solving modulus questions we take two cases 1) x>0 2) x<0. According to the first statement when x>0 we get x=9 which is valid, but when x<0 we get x=1 (which is not valid). Now, in some of the earlier questions when x<0 and if we got a positive value for it we neglected it and considered that x had only one valid value. In this question why hasn't something similar been done



Bunel, I too have the same doubt in my mind. I will try to explain it.

By taking condition,
X>0, i got the value of X as 9

And by taking the condition X<0, I have got the solution as X=1, Which is not a valid solution with the given condition.

So i took X=9 only and got the answer as A.


Pls explain how X=1 with the condition X<0 is valid.


Thanks in advance

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Re: Is root{x} a prime number? [#permalink]

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New post 27 Dec 2015, 04:31
ArunpriyanJ wrote:
mohnish104 wrote:
Bunuel wrote:

Is \(\sqrt{x}\) a prime number?

1) \(|3x-7|=2x+2\) --> we have one check point 7/3 (check point - the value of x for which an expression in absolute value equals to zero):

A. \(x\leq{\frac{7}{3}}\) --> \(3x-7\leq{0}\) hence \(|3x-7|=-(3x-7)\) --> \(-(3x-7)=2x+2\) --> \(x=1\) --> \(\sqrt{1}=1\neq{prime}\).
B. A. \(x>{\frac{7}{3}}\) --> \(3x-7>0\) hence \(|3x-7|=3x-7\) --> \(3x-7=2x+2\) --> \(x=9\) --> \(\sqrt{9}=3=prime\).
Two different answer. Not sufficient.

2) \(x^2=9x\) --> \(x(x-9)=0\) --> \(x=0\) or \(x=9\) --> \(\sqrt{0}=0\neq{prime}\) or \(\sqrt{9}=3=prime\). Not sufficient.

(1)+(2) Intersection of values from (1) and (2) is \(x=9\) --> \(\sqrt{9}=3=prime\). Sufficient.

Answer: C.





Bunuel, I have a question. Usually while solving modulus questions we take two cases 1) x>0 2) x<0. According to the first statement when x>0 we get x=9 which is valid, but when x<0 we get x=1 (which is not valid). Now, in some of the earlier questions when x<0 and if we got a positive value for it we neglected it and considered that x had only one valid value. In this question why hasn't something similar been done



Bunel, I too have the same doubt in my mind. I will try to explain it.

By taking condition,
X>0, i got the value of X as 9

And by taking the condition X<0, I have got the solution as X=1, Which is not a valid solution with the given condition.

So i took X=9 only and got the answer as A.


Pls explain how X=1 with the condition X<0 is valid.


Thanks in advance


Why are you talking about x<0 and x>0 when the transition point is 7/3, not 0?

You should brush up fundamentals on absolute value:

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Re: Is root{x} a prime number? [#permalink]

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New post 26 Jan 2016, 01:56
ArunpriyanJ wrote:
Bunuel, I have a question. Usually while solving modulus questions we take two cases 1) x>0 2) x<0. According to the first statement when x>0 we get x=9 which is valid, but when x<0 we get x=1 (which is not valid). Now, in some of the earlier questions when x<0 and if we got a positive value for it we neglected it and considered that x had only one valid value. In this question why hasn't something similar been done


Both x=1 and x=9 are valid for (1). Please elaborate what you mean?[/quote]

Bunel, I too have the same doubt in my mind. I will try to explain it.

By taking condition,
X>0, i got the value of X as 9

And by taking the condition X<0, I have got the solution as X=1, Which is not a valid solution with the given condition.

So i took X=9 only and got the answer as A.


Pls explain how X=1 with the condition X<0 is valid.


Thanks in advance[/quote]

we are taking condition x<=7/3 to be equation 3x-7 as <=0
So x=1 is acceptable

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Re: Is root{x} a prime number? [#permalink]

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New post 26 Jan 2016, 08:14
LM wrote:
Is \(\sqrt{x}\) a prime number?

(1) \(|3x-7|=2x+2\)

(2) \(x^2=9x\)


Fact (1) |3x-7|=2x+2
==> 3x-7=+(2x+2) or 3x-7=-(2x+2)
==> x=9 (\(\sqrt{9}=3\) YES) or x=1 (\(\sqrt{1}=1\) NO) Hence INSUFF.

Fact (2) x^2=9x
==> x^2-9x=0
==> x(x-9)=0
==> x=0 (\(\sqrt{0}=0\) NO) or x=9 (\(\sqrt{9}=3\) YES) Hence INSUFF.
(1)+(2) gives x=9 Hence SUFF
Ans: C
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Re: Is root{x} a prime number? [#permalink]

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Is root{x} a prime number? [#permalink]

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New post 05 Aug 2017, 02:39
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From 1 x=1 or 9. Hence sqrt(x) has 2 values. Not sufficient.
From 2 x=0 or 9. Hence sqrt(x) has 2 values. Not sufficient.
Both: x=9. Hence sqrt(x) is 3 which is prime. Sufficient.
Hence C. :)


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Re: Is root{x} a prime number? [#permalink]

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New post 05 Aug 2017, 02:57
LM wrote:
Is \(\sqrt{x}\) a prime number?

(1) \(|3x-7|=2x+2\)

(2) \(x^2=9x\)


1) |3x - 7| = 2x + 2
=> x = 1, 9
Insufficient.

2) \(x^2\) = 9x
=> x = 0,9
Insufficient.

1+2)
x = 0,1,9
=> x = 0, 1 do not support both statements so we cannot use them.
Only value of x = 9 remains.
=> \(\sqrt{x}\) = 3
Sufficient.

C is the answer.
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Re: Is root{x} a prime number? [#permalink]

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New post 08 Aug 2017, 04:39
x=1 and x=9 both are valid

|3x-7|=|3*1-7|=|-4|=4(modulus of any value is +ve)

2x+2=2*1+2=4.

Hence valid



ArunpriyanJ wrote:
Bunuel, I have a question. Usually while solving modulus questions we take two cases 1) x>0 2) x<0. According to the first statement when x>0 we get x=9 which is valid, but when x<0 we get x=1 (which is not valid). Now, in some of the earlier questions when x<0 and if we got a positive value for it we neglected it and considered that x had only one valid value. In this question why hasn't something similar been done


Both x=1 and x=9 are valid for (1). Please elaborate what you mean?[/quote]

Bunel, I too have the same doubt in my mind. I will try to explain it.

By taking condition,
X>0, i got the value of X as 9

And by taking the condition X<0, I have got the solution as X=1, Which is not a valid solution with the given condition.

So i took X=9 only and got the answer as A.


Pls explain how X=1 with the condition X<0 is valid.


Thanks in advance[/quote]

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Re: Is root{x} a prime number? [#permalink]

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New post 08 Aug 2017, 06:10
gps5441 wrote:
x=1 and x=9 both are valid

|3x-7|=|3*1-7|=|-4|=4(modulus of any value is +ve)

2x+2=2*1+2=4.

Hence valid



ArunpriyanJ wrote:
Bunuel, I have a question. Usually while solving modulus questions we take two cases 1) x>0 2) x<0. According to the first statement when x>0 we get x=9 which is valid, but when x<0 we get x=1 (which is not valid). Now, in some of the earlier questions when x<0 and if we got a positive value for it we neglected it and considered that x had only one valid value. In this question why hasn't something similar been done

Both x=1 and x=9 are valid for (1). Please elaborate what you mean?

Bunel, I too have the same doubt in my mind. I will try to explain it.

By taking condition,
X>0, i got the value of X as 9

And by taking the condition X<0, I have got the solution as X=1, Which is not a valid solution with the given condition.

So i took X=9 only and got the answer as A.


Pls explain how X=1 with the condition X<0 is valid.


Thanks in advance


Statement 1
X = 1 or 9 (Both the values satisfy the equation)
If x = 1, then \(\sqrt{x}\) = 1 => Not a prime number.
If x = 9, then \(\sqrt{x}\) = 3 => Prime Number
2 different answers, so this is insufficient.

Statement 2
x = 0 or 9 (Both the values satisfy the equation)
If x = 0, then \(\sqrt{x}\) = 0 => Not a prime number.
If x = 9, then \(\sqrt{x}\) = 3 => Prime Number
2 different answers, so this is insufficient.

Statement 1 + Statement 2
x = 9 (The only common value)
=> \(\sqrt{x}\) = 3 => Prime Number
Unique Answer and hence, Sufficient.
C is the answer.

Hope this helps!
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Re: Is root{x} a prime number?   [#permalink] 08 Aug 2017, 06:10

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