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Is root(x5)^2 = 5  x ? [#permalink]
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13 Jan 2007, 10:52
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Is \(\sqrt{(x5)^2} = 5  x\)? (1) xx > 0 (2) 5  x > 0
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For me (D)
sqrt( (x5)^2) = 5  x ?
<=> x5 = 5x ?
<=> 5x = 5x ?
This is true if 5x >= 0 <=> x =< 5
So we end up with checking if x =< 5 ?
Stat 1
x x > 0
<=> x < 0 < 5
SUFF.
Stat 2
5  x > 0
<=> x < 5
SUFF.



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Re: DS: Sqrt inequality [#permalink]
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15 Jun 2009, 11:25
My line of thinking is that the only time you can answer "no" to the question is when x>=5. For all values below 5 the answer is "yes"
1) tells us that x is negative  sufficient 2) tells us that x is <5  sufficient



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Re: DS: Sqrt inequality [#permalink]
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09 Sep 2009, 11:01
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Just want to make sure that my line of thinking is right. I got D as well.
sqrt(x5)^2 = 5x ==> x5 = 5x
which means that if
x5<0, (x5) = 5x = no solution, therefore x<5
x5>0, x5 = 5x, which means that x>5 and x = 5
from (1) xx > 0, we know that x has to be negative (x)x is the only way to get a number greater than 0. Therefore, this means that of the three possible solutions for x, only this x<5 hold true.
from (2) 5x>0, therefore x<5.
Can someone please point out if there is something wrong with my reasoning.



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Is root(x5)^2 = 5  x ? [#permalink]
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09 Sep 2009, 14:41
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dhushan wrote: Just want to make sure that my line of thinking is right. I got D as well.
sqrt(x5)^2 = 5x ==> x5 = 5x
which means that if
x5<0, (x5) = 5x = no solution, therefore x<5
x5>0, x5 = 5x, which means that x>5 and x = 5
from (1) xx > 0, we know that x has to be negative (x)x is the only way to get a number greater than 0. Therefore, this means that of the three possible solutions for x, only this x<5 hold true.
from (2) 5x>0, therefore x<5.
Can someone please point out if there is something wrong with my reasoning. Is \(\sqrt{(x5)^2}=5x\)?First of all, recall that \(\sqrt{x^2}=x\). Is \(\sqrt{(x5)^2}=5x\)? > is \(x5=5x\)? > is \(x5\leq{0}\)? > is \(x\leq{5}\)? (1) \(xx > 0\) > \(x\) is never negative (positive or zero), so for \(xx\) to be positive, \(x\) must be positive \(x>0\) > \(x<0\). Sufficient. (2) \(5x>0\) > \(x<5\). Sufficient. Answer: D.
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Re: Square root. [#permalink]
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29 Dec 2009, 11:27
For the question to be true. the right side of the equation has to be positive, (hence x has to be smaller than 5) because the left side of the equation is always positive.
s1 tells us that x is a negative number. So, it's sufficient s2 tells us that x is less than 5, so it sufficient.
Therefore, answer D



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Re: Square root. [#permalink]
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31 Dec 2009, 08:40
The question basically wants to know if x<=5 else RHS will be x5 Statement 1 xx>0 or xx<0 (multiply by 1 both sides and reverse the sign) either x<0 or x<0 since x is always positive or 0 x<0 is true. if x<0 then x<5 hence sufficient Statement 2 5x>0 5>x This is what we are looking for hence sufficient Answer is D.
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Re: Square root. [#permalink]
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31 Dec 2009, 08:49
Since L.H.S is a square it will be always positive.and R.H.S 5x will only be positive when x is less than or equal to 5. so in other words the question can be rephrase as is x<=5
stmt1: xx>0
divide both side by x we get x>0 it means x < 0 sufficient
stmt 2: 5x>0
add x both side we get 5>x means x<5 sufficient.
hope this helps



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Re: Square root. [#permalink]
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31 Dec 2009, 21:25
Simplify the question: x5=5x
statement 1: x<0
statement 2: 5>x
plug in any numbers for each statement. Hence, D



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Re: Square root. [#permalink]
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19 Mar 2011, 19:20
arjunrampal wrote: Please explain the approach to solve the problem and point to any relevant material available in the GMAT club. Attachment: square_root_D.JPG Let me point out something here: You cannot square both sides to get Is \((\sqrt{(x5)^2})^2 = (5x)^2\) ? People sometimes get confused here. Why can you not square it? It is a question similar to 'Is x = 5?' Can you square both sides here and change the question to 'Is \(x^2 = 25\)?' Please remember, they are not the same. x^2 can be 25 even if x is not 5 ( when x = 5, even then x^2 = 25). Only if it is given to you that x = 5, then you can say that x^2 = 25. You can rephrase the question in the following manner (and many more ways) Is \((\sqrt{(x5)^2}) = (5x)\) ? Is \(x5 = (5x)\) ? or Is \(5x = (5x)\)? We know that x = x only when x >= 0 So \(5x = (5x)\) only when 5  x >= 0 or when x <= 5 Stmnt 1: xx > 0 Since x is always positive (or zero), x must be positive too. So x must be negative. If x < 0, then x is obviously less than 5. Sufficient. Stmnt 2: 5  x> 0 x < 5. Sufficient Answer D
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Re: Square root. [#permalink]
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20 Mar 2011, 01:39
sqrt((x5)^2) = x5 If (x5) < 0 then x5 = 5 x So the question is if x < 5 From (1), xx > 0 => x is +ve, so x is ve hence sufficient. From (2), 5x > 0, so x5 < 0, which is what the question is asking, so answer is D.
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Re: Square root. [#permalink]
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20 Mar 2011, 12:39
sqrt(x^2) = x
x5 will only be equal to 5x when x5<0 => x has to be less than 5. so we have to see if x<5?
1. Sufficient
xx <0 x is always +ve => x is ve i.e x<0<5 = >x<5
2. Sufficient x<5
Answer D



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Re: Is root(x5)^2 = 5  x ? [#permalink]
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26 Jul 2014, 11:51
you need to prove that x<= 5 so x = 5 & x<5 Statement 2 doesnt prove x<5 Something I'm missing?
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Re: Is root(x5)^2 = 5  x ? [#permalink]
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26 Jul 2014, 12:00



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Is root(x5)^2 = 5  x ? [#permalink]
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11 Jul 2015, 01:04
I simplified the equation up to the point where it states: Is \((x4)*(x5) = 0\) or is x1 = 4 / x2 = 5 ?
I then was not able to use statement 1 properly, but would that theoretically be possible?



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Re: Is root(x5)^2 = 5  x ? [#permalink]
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11 Jul 2015, 01:29



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Re: Is root(x5)^2 = 5  x ? [#permalink]
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11 Jul 2015, 04:06
Bunuel wrote: noTh1ng wrote: I simplified the equation up to the point where it states: Is \((x4)*(x5) = 0\) or is x1 = 4 / x2 = 5 ?
I then was not able to use statement 1 properly, but would that theoretically be possible? It's not clear HOW you get the above. Please elaborate. I think i miscalculated. Now did it again, simplifying the equation \(\sqrt{(x5)^2} = 5  x\)? should result in 25=25, correct?



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Re: Is root(x5)^2 = 5  x ? [#permalink]
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11 Jul 2015, 07:40



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Re: Is root(x5)^2 = 5  x ? [#permalink]
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12 Jul 2015, 05:49
Bunuel wrote: dhushan wrote: Just want to make sure that my line of thinking is right. I got D as well.
sqrt(x5)^2 = 5x ==> x5 = 5x
which means that if
x5<0, (x5) = 5x = no solution, therefore x<5
x5>0, x5 = 5x, which means that x>5 and x = 5
from (1) xx > 0, we know that x has to be negative (x)x is the only way to get a number greater than 0. Therefore, this means that of the three possible solutions for x, only this x<5 hold true.
from (2) 5x>0, therefore x<5.
Can someone please point out if there is something wrong with my reasoning. Is \(\sqrt{(x5)^2}=5x\)?First of all, recall that \(\sqrt{x^2}=x\). Is \(\sqrt{(x5)^2}=5x\)? > is \(x5=5x\)? > is \(x5\leq{0}\)? > is \(x\leq{5}\)? (1) \(xx > 0\) > \(x\) is never negative (positive or zero), so for \(xx\) to be positive, \(x\) must be positive \(x>0\) > \(x<0\). Sufficient. (2) \(5x>0\) > \(x<5\). Sufficient. Answer: D. Hi Bunuel, I didn't understand from here : is x−5=5−x? > is x−5≤0? > is x≤5?. Can you please explain further ? Regards Kshitij



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Re: Is root(x5)^2 = 5  x ? [#permalink]
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12 Jul 2015, 07:32
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kshitij89Modulus always results in an positive value. If x5>0 then x−5 should be equal to x5 However, If x5<0 then modulus would result the positive value of it i.e. (x5)=5x Thus, is x−5=5−x? > is x−5<0? > is x<5? Solving as Bunuel ,you'll get D.




Re: Is root(x5)^2 = 5  x ?
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