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

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Is root(x-5)^2 = 5 - x ? [#permalink]  13 Jan 2007, 10:52
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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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[#permalink]  13 Jan 2007, 11:11
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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]  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]  09 Sep 2009, 11:01
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]  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(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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Re: Is root(x-5)^2 = 5 - x ? [#permalink]  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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Math Expert
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Re: Is root(x-5)^2 = 5 - x ? [#permalink]  26 Jul 2014, 12:00
Expert's post
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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Re: Is root(x-5)^2 = 5 - x ?   [#permalink] 26 Jul 2014, 12:00
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# Is root(x-5)^2 = 5 - x ?

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