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M08 #16 - squares

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M08 #16 - squares [#permalink] New post 01 Feb 2009, 07:09
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If each expression under the square root is greater than or equal to 0, what is \sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3?

(A) \sqrt{2-x}
(B) 2x-6 + \sqrt{2-x}
(C) \sqrt{2-x} + x-3
(D) 2x-6 + \sqrt{x-2}
(E) x + \sqrt{x-2}

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A

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Re: squares - m08,q16 [#permalink] New post 05 Feb 2009, 03:25
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A.

Since, each expression under the square root is positive, 2-x > 0 or, x < 2.

Thus, x-3 < 0.

Now, first expression = sqrt(x-3)^2 = |x-3| = 3-x

Thus, the whole expression will reduce to sqrt(2-x)
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Re: squares - m08,q16 [#permalink] New post 09 Feb 2009, 05:05
>>Now, first expression = sqrt(x-3)^2 = |x-3| = 3-x

Can you explain the above? Isn't the sqrt of a number always its positive root?

sqrt(x-3)^2 = x-3 ? NOT |x-3| ?
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Re: squares - m08,q16 [#permalink] New post 09 Feb 2009, 09:38
ConkergMat wrote:
>>Now, first expression = sqrt(x-3)^2 = |x-3| = 3-x

Can you explain the above? Isn't the sqrt of a number always its positive root?

sqrt(x-3)^2 = x-3 ? NOT |x-3| ?


You are right. That is why 3-x as (x-3) is negative.
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Re: squares - m08,q16 [#permalink] New post 17 Oct 2009, 17:43
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xALIx wrote:
scthakur wrote:
ConkergMat wrote:
>>Now, first expression = sqrt(x-3)^2 = |x-3| = 3-x

Can you explain the above? Isn't the sqrt of a number always its positive root?

sqrt(x-3)^2 = x-3 ? NOT |x-3| ?


You are right. That is why 3-x as (x-3) is negative.


I still don't get how you got 3-x.
Can someone please explain!!!.


It's way too late for the reply but here it goes

x< 2 and sqrt ( x-3 ) ^2 >=0

now for sqrt(x-3)^2 there are two values : x-3 or -(x-3)
x-3>0 and x>3 - This is impossible because we have x<2

So sqrt(x-3)^2 = 3-x.
The rest is pretty straightforward.
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Re: M08 #16 - squares [#permalink] New post 16 Aug 2010, 06:14
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I think it should be B.

sqrt((x-3) (x-3)) should give (x-3) because it is mentioned that each expression under the square root is greater than or equal to 0.

Rest is simple --> x-3 +sqrt(x-2) +x -3

2x -6 +sqrt(x-2)
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Re: M08 #16 - squares [#permalink] New post 16 Aug 2010, 07:49
2-x> o then x<2 or x is negative

(x-3)^2 >o then x is different from 3

if x is negative then squareroot((x-3)^2 )=|x-3|=3-x then answer a

if 0<x<2 with x different from 3 then squareroot((x-3)^2 )=|x-3|=3-x then answer a too

in both cases x-3 is negative hence the squareroot of it's square is equal to |x-3|

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Re: M08 #16 - squares [#permalink] New post 16 Aug 2010, 08:46
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ConkergMat wrote:
If each expression under the square root is greater than or equal to 0, what is \sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3?

(A) \sqrt{2-x}
(B) 2x-6 + \sqrt{2-x}
(C) \sqrt{2-x} + x-3
(D) 2x-6 + \sqrt{x-2}
(E) x + \sqrt{x-2}

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A

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One important note: \sqrt{x^2}=|x|

\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3.

Now, as the expressions under the square roots are more than or equal to zero than 2-x>0 --> x<2. Next: as x<2 then |x-3| becomes |x-3|=-(x-3)=-x+3.

|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}.

Answer: A.
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Re: M08 #16 - squares [#permalink] New post 16 Aug 2010, 09:57
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I have done it as follows..

Plugin a value for x where the value of the expression under the squareroot will be greater than or equal to zero, i.e., x^2-6x+9 >=0 and 2-x >=0.

I took x=1, substituing that gives 4 for 1st expression and 1 for 2nd expression. Now, Substituing x=1 in the answer choices. Except option A, all other choices give a negative answer. So, answer is A.
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Re: M08 #16 - squares [#permalink] New post 17 Aug 2010, 00:41
Its A.

Since each expression under the square root is greater than or equal to 0, x<2. For all values of x< 1, the equation is true.

Lets suppose x = 1! Putting x=1 in sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3 gives 1 as solution.

Only answer choice 1 gives the same answer.
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Re: M08 #16 - squares [#permalink] New post 17 Aug 2010, 07:14
a is for me

\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3 = |x-3|+ \sqrt{2 - x} + x - 3

because 2-x >=0, x<= 2, |x-3| can be expressed as 3-x

the result is 3-x + \sqrt{2 - x} + x - 3 = \sqrt{2 - x}
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Re: M08 #16 - squares [#permalink] New post 13 May 2011, 02:25
ssing1 wrote:
I think it should be B.

sqrt((x-3) (x-3)) should give (x-3) because it is mentioned that each expression under the square root is greater than or equal to 0.

Rest is simple --> x-3 +sqrt(x-2) +x -3

2x -6 +sqrt(x-2)

nope
it is mentioned that the expression under the square root > 0.
Also square root of 16 is 4 and not 4,-4.
There4, sgrt [(x-3)(x-3) ] becomes 3-x and not x-3 to keep the sqrt positive since x<2.
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Re: M08 #16 - squares [#permalink] New post 01 Jul 2011, 07:24
Awesome question! I fell for B like a mortal! Every piece of information on the GMAT counts, if the question stem mentions that every expression under the square is greater than or equal to zero, we HAVE to use it somewhere!
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Re: M08 #16 - squares [#permalink] New post 07 Jul 2011, 10:23
AriBenCanaan wrote:
Awesome question! I fell for B like a mortal! Every piece of information on the GMAT counts, if the question stem mentions that every expression under the square is greater than or equal to zero, we HAVE to use it somewhere!


agree, probably questions on gmat may not be so straight forward after all :idea:
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Re: M08 #16 - squares [#permalink] New post 12 Feb 2012, 03:45
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Pretty tough....its not the question itself tough, but remembering underlying algebra principles...

whenever I see the expression a^2 -2.a.b + b^2 I immediately think it as (a-b)^2, as we learned in the school. But the fact is, it depends on how one perceives.
It can be either (a-b)^2 or (b-a)^2 as there is no info about a and b. That is why we should remember the value of \sqrt{a^2 -2.a.b + b^2} as |a-b| and it should be evaluated based on the a & b variables info. This is what actually tested in the question - a tough question, but very good question.
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Re: M08 #16 - squares [#permalink] New post 20 Aug 2012, 09:09
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ConkergMat wrote:
If each expression under the square root is greater than or equal to 0, what is \sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3?

(A) \sqrt{2-x}
(B) 2x-6 + \sqrt{2-x}
(C) \sqrt{2-x} + x-3
(D) 2x-6 + \sqrt{x-2}
(E) x + \sqrt{x-2}

[Reveal] Spoiler: OA
A

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Suppose choose the value x=1 which satisfies the condition
Hence the given becomes 2+1+1-3=1
1.1
2.3.4.5.not equal
Hence 1 is correct
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Re: M08 #16 - squares [#permalink] New post 29 Aug 2012, 22:40
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Really good question. There are a few parts to this question and I did all but one correct. I could make out that the first expression is (x-3)^2 and that there are two options i.e. +(x-3) and -(x-3). Using these two values I got A and B as the possible answer choices.

At this point the question stem that each expression under the square root is greater than or equal to zero.

This is where I made the mistake and chose B. Looking back through the answer choice right now it was really a stupid mistake to consider the option as -(x-3). I misinterpreted the question stem of value under the square root as non-negative for the value of x to be non-negative. A real stupid blunder.

Really good question though. Nice learning.
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Re: M08 #16 - squares [#permalink] New post 22 Aug 2013, 03:38
Yep, answer is A.

I spent 3.41 on this one, after looking at it for about a minute... easy solution is to substitute a value of '1' into the expression (obtaining '1' as a result), then confirm which answer option provides the same value.

Only option A also produces '1': SQRT(2-1) = 1.

Fairly easy way to go about this one, just takes up some time (WORTH IT TO GET THE CORRECT ANSWER)
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Re: M08 #16 - squares [#permalink] New post 20 Oct 2013, 13:30
Hi Bunuel,
You posted: as x<2, then |x-3|=-(x-3)=-x+3

But |x-3| equals either x-3 or -(x-3)
If x-3>0, then x>3
If x-3<0, then x-3<-0, x<3

How did you conclude to use -(x-3) for |x-3|?

Thanks,
Francis
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Re: M08 #16 - squares [#permalink] New post 20 Oct 2013, 13:39
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jrawls wrote:
Hi Bunuel,
You posted: as x<2, then |x-3|=-(x-3)=-x+3

But |x-3| equals either x-3 or -(x-3)
If x-3>0, then x>3
If x-3<0, then x-3<-0, x<3

How did you conclude to use -(x-3) for |x-3|?

Thanks,
Francis


ABSOLUTE VALUE PROPERTIES:

When x\leq{0} then |x|=-x, or more generally when some \ expression\leq{0} then |some \ expression|={-(some \ expression)}. For example: |-5|=5=-(-5);

When x\geq{0} then |x|=x, or more generally when some \ expression\geq{0} then |some \ expression|={some \ expression}. For example: |5|=5.

BACK TO THE QUESTION:
|x-3|=x-3, when x>3.
|x-3|=-(x-3), when x<3.

Since we know that x<2, then we have the second case (when x<2, x-3<0, thus |x-3|=-(x-3)).

Hope it's clear.
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Re: M08 #16 - squares   [#permalink] 20 Oct 2013, 13:39
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