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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 \(|x-3|=-(x-3)=-x+3\).
Re: M08 #16 - squares [#permalink]
16 Aug 2010, 09:57
1
This post received KUDOS
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.
Re: M08 #16 - squares [#permalink]
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.
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!
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
Re: M08 #16 - squares [#permalink]
12 Feb 2012, 03:45
1
This post received KUDOS
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. Thanks _________________
Re: M08 #16 - squares [#permalink]
29 Aug 2012, 22:40
1
This post received KUDOS
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. _________________
My attempt to capture my B-School Journey in a Blog : tranquilnomadgmat.blogspot.com
Re: M08 #16 - squares [#permalink]
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)
Re: M08 #16 - squares [#permalink]
20 Oct 2013, 13:39
2
This post received KUDOS
Expert's post
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)).