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Is root(x5)^2 = 5  x ?
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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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Is root(x5)^2 = 5  x ?
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09 Sep 2009, 14:41
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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Is sqrt(x5)^2 = 5x?
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06 Sep 2010, 11:20
Is \(\sqrt {(x5)^2} = 5x\) ? (1) xx > 0 (2) 5  x > 0 PS. Is always this true?: \(\sqrt{x^2}\) = lxl ?
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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
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09 Sep 2009, 11:01
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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Re: Square root.
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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: Is sqrt(x5)^2 = 5x?
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06 Sep 2010, 11:30
metallicafan wrote: Is \(sqrt(x5)^2 = 5x?\)
(1) xlxl > 0 (2) 5x > 0
PS. Is always this true?: \(sqrt{x^2}\) = lxl ? Is \(\sqrt{(x5)^2}=5x\)? Remember: \(\sqrt{x^2}=x\). So "is \(\sqrt{(x5)^2}=5x\)?" becomes: is \(x5=5x\)? \(x5=5x\) is true only for \(x\leq{5}\), as in this case \(\{LHS=x5=5x\}=\{RHS=5x\}\). So we have that if \(x\leq{5}\), then \(x5=5x\) is true. Basically question asks is \(x\leq{5}\)? (1) \(xx > 0\) > \(x\) is never negative (positive or zero), so in order to have \(xx > 0\), \(x\) must be positive \(x>0\) > \(x<0\), so \(x\) is less than 5 too. Sufficient. (2) \(5x>0\) > \(x<5\). Sufficient. Answer: D. Hope it helps. P.S. Explanation of: \(\sqrt{x^2}=x\). The point here is that as square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\). So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to? Let's consider following examples: If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\); If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\). So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=x\), if \(x<0\). What function does exactly the same thing? The absolute value function: \(x=x\), if \(x\geq{0}\) and \(x=x\), if \(x<0\). That is why \(\sqrt{x^2}=x\).
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Re: Is sqrt(x5)^2 = 5x?
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07 Sep 2010, 01:44
Bunuel wrote: metallicafan wrote: Is \(sqrt(x5)^2 = 5x?\)
(1) xlxl > 0 (2) 5x > 0
PS. Is always this true?: \(sqrt{x^2}\) = lxl ? Is \(\sqrt{(x5)^2}=5x\)? Remember: \(\sqrt{x^2}=x\). So "is \(\sqrt{(x5)^2}=5x\)?" becomes: is \(x5=5x\)? \(x5=5x\) is true only for \(x\leq{5}\), as in this case \(\{LHS=x5=5x\}=\{RHS=5x\}\). So we have that if \(x\leq{5}\), then \(x5=5x\) is true. Basically question asks is \(x\leq{5}\)? (1) \(xx > 0\) > \(x\) is never negative (positive or zero), so in order to have \(xx > 0\), \(x\) must be positive \(x>0\) > \(x<0\), so \(x\) is less than 5 too. Sufficient. (2) \(5x>0\) > \(x<5\). Sufficient. Answer: D. Hope it helps. Hi Bunuel, I get confused in the following concept. Can you please help me: \(xx > 0\) > \(x\) is never negative (positive or zero), so in order to have \(xx > 0\), \(x\) must be positive \(x>0\) > \(x<0\), so \(x\) is less than 5 too Can't we solve it as follows: As \(x\) is never negative > \(xx > 0\) = x*x = x^2 >0 = x^2<0 (multiplying by ve sign and flipping sign) x^2<0 => \(sqrt{x^2}\) <0 => lxl <0 (as \(sqrt{x^2}\) =lxl ) Since lxl cannot be negative and lxl <0 that implies X<0 I have reached to same conclusion as yours but wanted to confirm if my approach is right. Please explain.Thanks
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Re: Is sqrt(x5)^2 = 5x?
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07 Sep 2010, 04:01
oldstudent wrote: Bunuel wrote: metallicafan wrote: Is \(sqrt(x5)^2 = 5x?\)
(1) xlxl > 0 (2) 5x > 0
PS. Is always this true?: \(sqrt{x^2}\) = lxl ? Is \(\sqrt{(x5)^2}=5x\)? Remember: \(\sqrt{x^2}=x\). So "is \(\sqrt{(x5)^2}=5x\)?" becomes: is \(x5=5x\)? \(x5=5x\) is true only for \(x\leq{5}\), as in this case \(\{LHS=x5=5x\}=\{RHS=5x\}\). So we have that if \(x\leq{5}\), then \(x5=5x\) is true. Basically question asks is \(x\leq{5}\)? (1) \(xx > 0\) > \(x\) is never negative (positive or zero), so in order to have \(xx > 0\), \(x\) must be positive \(x>0\) > \(x<0\), so \(x\) is less than 5 too. Sufficient. (2) \(5x>0\) > \(x<5\). Sufficient. Answer: D. Hope it helps. Hi Bunuel, I get confused in the following concept. Can you please help me: \(xx > 0\) > \(x\) is never negative (positive or zero), so in order to have \(xx > 0\), \(x\) must be positive \(x>0\) > \(x<0\), so \(x\) is less than 5 too Can't we solve it as follows: As \(x\) is never negative > \(xx > 0\)= x*x = x^2 >0 = x^2<0 (multiplying by ve sign and flipping sign) x^2<0 => \(sqrt{x^2}\) <0=> lxl <0 (as \(sqrt{x^2}\) =lxl ) Since lxl cannot be negative and lxl <0 that implies X<0 I have reached to same conclusion as yours but wanted to confirm if my approach is right. Please explain.Thanks This approach is not right. The red parts are not correct. \(xx > 0\) can not be written as \(x*x>0\), as \(x\geq{0}\) does not mean \(x\) itself can not be negative > \(x=x\) if \(x\geq{0}\) and \(x=x\) if \(x\leq{0}\), so when you are writing \(x\) instead of \(x\) you are basically assuming that \(x\geq{0}\) and then in the end get the opposite result \(x<0\). Next, \(x^2<0\) has no solution, square of a number can not be negative, so no \(x\) can make this inequality hold true.
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Re: Is sqrt(x5)^2 = 5x?
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07 Sep 2010, 22:42
I think for this question we dont need any statement... without statement itself it is possible to say if the equality is correct or wrong. Can someone comment on this... Bunuel please?



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Re: Is sqrt(x5)^2 = 5x?
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Re: Square root.
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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: Is root(x5)^2 = 5  x ?
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12 Jul 2015, 07:32
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.



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Is root(x5)^2 = 5  x ?
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07 Dec 2018, 20:01
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. in the highlighted portion, why we can not also write that x is greater than or equal to 5. if we take out modulus, x5 can be positive and can be negative. there are two possible values of x. you consider only negative possibility. please help. i don't understand.
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Re: Is root(x5)^2 = 5  x ?
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07 Dec 2018, 21:18
rashedBhai wrote: 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. in the highlighted portion, why we can not also write that x is greater than or equal to 5. if we take out modulus, x5 can be positive and can be negative. there are two possible values of x. you consider only negative possibility. please help. i don't understand. Hello I will try to explain. If x is >= 5 (greater than or equal to 5), then x5 will be positive, and square root of (x5)^2, will be equal to x5 only, it will never be equal to 5x. Eg, lets take x=6, here x5 = 1, which is positive. Here (x5)^2 = 1^2 = 1, and its square root is 1, which is x5 only, NOT 5x. But if we take x=4, here x5 = 1, which is negative. Here (x5)^2 = (1)^2 = 1, and its square root is 1, which is NOT x5: rather its 5x.




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