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

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Is root(x-5)^2 = 5 - x ? [#permalink] New post 13 Jan 2007, 10:52
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Is \(\sqrt{(x-5)^2} = 5 - x\)?

(1) -x|x| > 0
(2) 5 - x > 0
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 [#permalink] New post 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] New post 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] New post 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] New post 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] New post 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(x-5)^2 = 5 - x ? [#permalink] New post 26 Jul 2014, 12:00
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Is root(x-5)^2 = 5 - x ? [#permalink] New post 11 Jul 2015, 01:04
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] New post 11 Jul 2015, 01:29
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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] New post 11 Jul 2015, 04:06
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] New post 11 Jul 2015, 07:40
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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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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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



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
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Re: Is root(x-5)^2 = 5 - x ? [#permalink] New post 12 Jul 2015, 07:32
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, 07:32
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