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# Is root{x} a prime number?

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Manager
Joined: 03 Aug 2015
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Re: Is root{x} a prime number? [#permalink]

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24 Dec 2015, 23:29
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 $$|3x-7|=3x-7$$ --> $$3x-7=2x+2$$ --> $$x=9$$ --> $$\sqrt{9}=3=prime$$.

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.

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.

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

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27 Dec 2015, 03:31
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 $$|3x-7|=3x-7$$ --> $$3x-7=2x+2$$ --> $$x=9$$ --> $$\sqrt{9}=3=prime$$.

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.

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.

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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26 Jan 2016, 00:56
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.

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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26 Jan 2016, 07:14
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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05 Aug 2017, 01:39
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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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05 Aug 2017, 01:57
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.

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

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08 Aug 2017, 03:39
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.

Kudos [?]: 21 [0], given: 126

Senior Manager
Joined: 06 Jul 2016
Posts: 437

Kudos [?]: 132 [0], given: 98

Location: Singapore
Concentration: Strategy, Finance
Re: Is root{x} a prime number? [#permalink]

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08 Aug 2017, 05:10
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.

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

Hope this helps!
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Re: Is root{x} a prime number?   [#permalink] 08 Aug 2017, 05:10

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