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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
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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.

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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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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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Is root(x-5)^2 = 5 - x ? [#permalink]  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]  11 Jul 2015, 01:29
Expert's post
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]  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]  11 Jul 2015, 07:40
Expert's post
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]  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.

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]  12 Jul 2015, 07:32
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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 ? [#permalink]  09 Oct 2015, 06:40
Hi Bunuel,

Could you please explain this step |x−5|=5−x? --> is x−5≤0? --> is x≤5?. I know for all values less than or equal to 5 holds good for this equation, however, how do we arrive to that final equation using |x−5|=5−x.

What i do understand is that absolute value of anything has two values. When i simplify the equation above i get x=5 (when i take positive value) and x-5=x-5 (when i take negative value).

Thanks.
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Re: Is root(x-5)^2 = 5 - x ? [#permalink]  09 Oct 2015, 06:51
Swaroopdev wrote:
Hi Bunuel,

Could you please explain this step |x−5|=5−x? --> is x−5≤0? --> is x≤5?. I know for all values less than or equal to 5 holds good for this equation, however, how do we arrive to that final equation using |x−5|=5−x.

What i do understand is that absolute value of anything has two values. When i simplify the equation above i get x=5 (when i take positive value) and x-5=x-5 (when i take negative value).

Thanks.

You know that $$\sqrt{x^2} = |x|$$ ---> $$\sqrt{x^2} = \pm x$$. Thus $$\sqrt{(x-5)^2} = |x-5|$$ --->$$\sqrt{(x-5)^2} = \pm (x-5)$$....(1)

Additionally, you know that $$\sqrt {x} \geq 0$$ (for GMAT purposes!)....(2)

Thus based on (1) and (2) above,

Coming back to your question, the questions asks whether $$\sqrt{(x-5)^2} = 5-x$$. This can only be possible when $$x\leq 5$$ as if x>5 then 5-x will become <0 and will go against (2) above. The most x can go is = 5....(3)

Also, from (1), $$\sqrt{x^2} = - x$$ only when x<0 ...(4)

Thus from (3) and (4), you get the rephrase of the question asked as is $$x \leq 5$$ ?

Hope this helps.
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Re: Is root(x-5)^2 = 5 - x ?   [#permalink] 09 Oct 2015, 06:51
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