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

(1) -x|x| > 0
(2) 5 - x > 0
[Reveal] Spoiler: OA
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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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15 Jun 2009, 12:25
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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09 Sep 2009, 12:01
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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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Is root(x-5)^2 = 5 - x ? [#permalink]

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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.

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29 Dec 2009, 12:27
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.

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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

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31 Dec 2009, 09:49
Since L.H.S is a square it will be always positive.and R.H.S 5-x 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: -x|x|>0

divide both side by |x| we get -x>0 it means x < 0 sufficient

stmt 2: 5-x>0

we get 5>x means x<5 sufficient.

hope this helps
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31 Dec 2009, 22:25
Simplify the question:
|x-5|=5-x

statement 1:
x<0

statement 2:
5>x

plug in any numbers for each statement.
Hence, D
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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

[Reveal] Spoiler:
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

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20 Mar 2011, 02:39
sqrt((x-5)^2) = |x-5|

If (x-5) < 0 then |x-5| = 5 -x

So the question is if x < 5

From (1), -x|x| > 0 => -x is +ve, so x is -ve hence sufficient.

From (2), 5-x > 0, so x-5 < 0, which is what the question is asking, so answer is D.
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20 Mar 2011, 13:39
sqrt(x^2) = |x|

|x-5| will only be equal to 5-x when x-5<0 => x has to be less than 5. so we have to see if x<5?

1. Sufficient

-x|x| <0
|x| is always +ve => x is -ve i.e x<0<5 = >x<5

2. Sufficient
x<5

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

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26 Jul 2014, 12:51
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(x-5)^2 = 5 - x ? [#permalink]

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26 Jul 2014, 13:00
amenon55 wrote:
you need to prove that
x<= 5 so x = 5 & x<5

Statement 2 doesnt prove x<5

Something I'm missing?

The question asks whether $$x\leq{5}$$. (2) states that $$x<5$$, so the answer to the question is YES.
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Is root(x-5)^2 = 5 - x ? [#permalink]

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11 Jul 2015, 02:04
I simplified the equation up to the point where it states:
Is $$(x-4)*(x-5) = 0$$ 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(x-5)^2 = 5 - x ? [#permalink]

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11 Jul 2015, 02:29
noTh1ng wrote:
I simplified the equation up to the point where it states:
Is $$(x-4)*(x-5) = 0$$ 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.
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Re: Is root(x-5)^2 = 5 - x ? [#permalink]

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11 Jul 2015, 05:06
Bunuel wrote:
noTh1ng wrote:
I simplified the equation up to the point where it states:
Is $$(x-4)*(x-5) = 0$$ 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 $$\sqrt{(x-5)^2} = 5 - x$$?

should result in 25=25, correct?
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Re: Is root(x-5)^2 = 5 - x ? [#permalink]

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11 Jul 2015, 08:40
noTh1ng wrote:
Bunuel wrote:
noTh1ng wrote:
I simplified the equation up to the point where it states:
Is $$(x-4)*(x-5) = 0$$ 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 $$\sqrt{(x-5)^2} = 5 - x$$?

should result in 25=25, correct?

No. The above equation holds for ANY value of x less than or equal to 5.
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Re: Is root(x-5)^2 = 5 - x ? [#permalink]

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12 Jul 2015, 06:49
Bunuel wrote:
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.

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(x-5)^2 = 5 - x ? [#permalink]

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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.
Re: Is root(x-5)^2 = 5 - x ?   [#permalink] 12 Jul 2015, 08:32

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