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Is root(x-5)^2 = 5 - x ?

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Is root(x-5)^2 = 5 - x ?  [#permalink]

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New post 13 Jan 2007, 11:52
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Is \(\sqrt{(x-5)^2} = 5 - x\)?

(1) -x|x| > 0
(2) 5 - x > 0
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Is root(x-5)^2 = 5 - x ?  [#permalink]

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New post 09 Sep 2009, 15:41
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dhushan wrote:
Just want to make sure that my line of thinking is right. I got D as well.

sqrt(x-5)^2 = 5-x
==> |x-5| = 5-x

which means that if

x-5<0, -(x-5) = 5-x = no solution, therefore x<5

x-5>0, x-5 = 5-x, which means that x>5 and x = 5

from (1) -x|x| > 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) 5-x>0, therefore x<5.

Can someone please point out if there is something wrong with my reasoning.


Is \(\sqrt{(x-5)^2}=5-x\)?

First of all, recall that \(\sqrt{x^2}=|x|\).

Is \(\sqrt{(x-5)^2}=5-x\)? --> is \(|x-5|=5-x\)? --> is \(x-5\leq{0}\)? --> is \(x\leq{5}\)?

(1) \(-x|x| > 0\) --> \(|x|\) is never negative (positive or zero), so for \(-x|x|\) to be positive, \(-x\) must be positive \(-x>0\) --> \(x<0\). Sufficient.

(2) \(5-x>0\) --> \(x<5\). Sufficient.

Answer: D.
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Is sqrt(x-5)^2 = 5-x?  [#permalink]

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New post 06 Sep 2010, 12:20
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Is \(\sqrt {(x-5)^2} = 5-x\) ?

(1) -x|x| > 0
(2) 5 - x > 0

PS. Is always this true?: \(\sqrt{x^2}\) = lxl ?
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  [#permalink]

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New post 13 Jan 2007, 12:11
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For me (D) :)

sqrt( (x-5)^2) = 5 - x ?
<=> |x-5| = 5-x ?
<=> |5-x| = 5-x ?

This is true if 5-x >= 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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New post 09 Sep 2009, 12:01
1
Just want to make sure that my line of thinking is right. I got D as well.

sqrt(x-5)^2 = 5-x
==> |x-5| = 5-x

which means that if

x-5<0, -(x-5) = 5-x = no solution, therefore x<5

x-5>0, x-5 = 5-x, which means that x>5 and x = 5

from (1) -x|x| > 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) 5-x>0, therefore x<5.

Can someone please point out if there is something wrong with my reasoning.
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Re: Square root.  [#permalink]

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New post 31 Dec 2009, 09:40
The question basically wants to know if x<=5 else RHS will be x-5

Statement 1
-x|x|>0
or
x|x|<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
5-x>0
5>x
This is what we are looking for hence sufficient

Answer is D.
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Re: Is sqrt(x-5)^2 = 5-x?  [#permalink]

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New post 06 Sep 2010, 12:30
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metallicafan wrote:
Is \(sqrt(x-5)^2 = 5-x?\)

(1) -xlxl > 0
(2) 5-x > 0

PS. Is always this true?: \(sqrt{x^2}\) = lxl ?


Is \(\sqrt{(x-5)^2}=5-x\)?

Remember: \(\sqrt{x^2}=|x|\).

So "is \(\sqrt{(x-5)^2}=5-x\)?" becomes: is \(|x-5|=5-x\)?

\(|x-5|=5-x\) is true only for \(x\leq{5}\), as in this case \(\{LHS=|x-5|=5-x\}=\{RHS=5-x\}\). So we have that if \(x\leq{5}\), then \(|x-5|=5-x\) is true.

Basically question asks is \(x\leq{5}\)?

(1) \(-x|x| > 0\) --> \(|x|\) is never negative (positive or zero), so in order to have \(-x|x| > 0\), \(-x\) must be positive \(-x>0\) --> \(x<0\), so \(x\) is less than 5 too. Sufficient.

(2) \(5-x>0\) --> \(x<5\). 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(x-5)^2 = 5-x?  [#permalink]

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New post 07 Sep 2010, 02:44
Bunuel wrote:
metallicafan wrote:
Is \(sqrt(x-5)^2 = 5-x?\)

(1) -xlxl > 0
(2) 5-x > 0

PS. Is always this true?: \(sqrt{x^2}\) = lxl ?


Is \(\sqrt{(x-5)^2}=5-x\)?

Remember: \(\sqrt{x^2}=|x|\).

So "is \(\sqrt{(x-5)^2}=5-x\)?" becomes: is \(|x-5|=5-x\)?


\(|x-5|=5-x\) is true only for \(x\leq{5}\), as in this case \(\{LHS=|x-5|=5-x\}=\{RHS=5-x\}\). So we have that if \(x\leq{5}\), then \(|x-5|=5-x\) is true.

Basically question asks is \(x\leq{5}\)?

(1) \(-x|x| > 0\) --> \(|x|\) is never negative (positive or zero), so in order to have \(-x|x| > 0\), \(-x\) must be positive \(-x>0\) --> \(x<0\), so \(x\) is less than 5 too. Sufficient.

(2) \(5-x>0\) --> \(x<5\). Sufficient.

Answer: D.

Hope it helps.



Hi Bunuel,
I get confused in the following concept. Can you please help me:


\(-x|x| > 0\) --> \(|x|\) is never negative (positive or zero), so in order to have \(-x|x| > 0\), \(-x\) 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 -> \(-x|x| > 0\) = -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(x-5)^2 = 5-x?  [#permalink]

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New post 07 Sep 2010, 05:01
oldstudent wrote:
Bunuel wrote:
metallicafan wrote:
Is \(sqrt(x-5)^2 = 5-x?\)

(1) -xlxl > 0
(2) 5-x > 0

PS. Is always this true?: \(sqrt{x^2}\) = lxl ?


Is \(\sqrt{(x-5)^2}=5-x\)?

Remember: \(\sqrt{x^2}=|x|\).

So "is \(\sqrt{(x-5)^2}=5-x\)?" becomes: is \(|x-5|=5-x\)?


\(|x-5|=5-x\) is true only for \(x\leq{5}\), as in this case \(\{LHS=|x-5|=5-x\}=\{RHS=5-x\}\). So we have that if \(x\leq{5}\), then \(|x-5|=5-x\) is true.

Basically question asks is \(x\leq{5}\)?

(1) \(-x|x| > 0\) --> \(|x|\) is never negative (positive or zero), so in order to have \(-x|x| > 0\), \(-x\) must be positive \(-x>0\) --> \(x<0\), so \(x\) is less than 5 too. Sufficient.

(2) \(5-x>0\) --> \(x<5\). Sufficient.

Answer: D.

Hope it helps.



Hi Bunuel,
I get confused in the following concept. Can you please help me:


\(-x|x| > 0\) --> \(|x|\) is never negative (positive or zero), so in order to have \(-x|x| > 0\), \(-x\) 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 -> \(-x|x| > 0\)= -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.

\(-x|x| > 0\) 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(x-5)^2 = 5-x?  [#permalink]

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New post 07 Sep 2010, 23:42
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(x-5)^2 = 5-x?  [#permalink]

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New post 08 Sep 2010, 06:59
amitjash wrote:
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?


This not correct. \(x\) must be less than or equal to 5 inequality \(\sqrt{(x-5)^2}=5-x\) to hold true. For example if \(x=10>5\) then \(\sqrt{(x-5)^2}=5\neq{-5}=5-x\)
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Re: Square root.  [#permalink]

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New post 19 Mar 2011, 20:20
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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


AO=D


Let me point out something here: You cannot square both sides to get
Is \((\sqrt{(x-5)^2})^2 = (5-x)^2\) ?

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 \((\sqrt{(x-5)^2}) = (5-x)\) ?
Is \(|x-5| = (5-x)\) ?
or Is \(|5-x| = (5-x)\)?
We know that |x| = x only when x >= 0
So \(|5-x| = (5-x)\) only when 5 - x >= 0 or when x <= 5

Stmnt 1: -x|x| > 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(x-5)^2 = 5 - x ?  [#permalink]

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New post 12 Jul 2015, 08:32
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kshitij89

Modulus always results in an positive value.
If x-5>0 then |x−5| should be equal to x-5
However, If x-5<0 then modulus would result the positive value of it i.e. -(x-5)=5-x

Thus, is |x−5|=5−x? --> is x−5<0? --> is x<5?

Solving as Bunuel ,you'll get D. :)
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Re: Is root(x-5)^2 = 5 - x ? &nbs [#permalink] 12 Jul 2015, 08:32
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