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If x+sqrt(x^2-4x+4)=2, then:

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GMATH Teacher
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Joined: 12 Oct 2010
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Updated on: 27 Mar 2019, 12:52
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95% (hard)

Question Stats:

19% (01:35) correct 81% (01:56) wrong based on 59 sessions

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GMATH practice exercise (Quant Class 12)

The alternative choices are tricky... the official answer is NOT wrong, and you can find it in 3 SECONDS, using LOGIC only!

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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net

Originally posted by fskilnik on 27 Mar 2019, 12:44.
Last edited by fskilnik on 27 Mar 2019, 12:52, edited 1 time in total.
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27 Mar 2019, 12:56
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Sent from my iPhone using GMAT Club Forum

Excellent! I have edited my "hint" to mention the possibility of finding the right answer in less than a few seconds. Congrats!

P.S.: if we take into account that GMAT asks the candidate to find the BEST answer, someone could argue that it is very "risky" to adopt the 3-seconds approach... think about that!
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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
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27 Mar 2019, 13:10
Doing this in the mathematical way (the 3-second answer just requires you to notice that the interval answer A contains the interval in every other answer choice, so if one of the other answers were correct, A would naturally also need to be correct, and a GMAT problem cannot have two different correct answers) --

x^2 - 4x + 4 = (x-2)^2, so √(x^2 - 4x + 4) = √(x-2)^2

Many test takers, seeing something like √(a^2), will think that is equal to a. That's one of the more common errors test takers make on higher level GMAT algebra questions. It is true that √(a^2) = a when a is positive or zero. But it's not true if a is negative -- if you plug a = -3 into √(a^2), you'll see that it is equal to 3, so the sign has flipped from negative to positive. In general, all we can say is that √(a^2) = |a| is always true.

So √(x-2)^2 = |x-2|, and our equation here becomes:

x + |x - 2| = 2

We can solve this using cases:

- if the thing inside the absolute value, x-2, is positive or zero, the absolute value will do nothing, so we can erase it and solve. Then we get

x + x - 2 = 2
2x = 4
x = 2

So when x - 2 > 0, there is only one solution, x = 2.

- when x-2 is negative, though, the absolute value will flip it's sign, so it will become 2-x. So when x < 2, our equation becomes

x + 2 - x = 2
2 = 2

and the equation is always true for every value of x < 2.

Combining our solutions from the two cases, the equation will be true whenever x < 2. If that were an answer choice, it would be the right answer, but we don't find that among the choices. But if we know for certain that x < 2, then it's clearly true that x < 3, so that's the right answer here. We can't be certain any of the other answer choices are correct, because some potential values of x lie outside the intervals in every other choice.

Nice question!
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27 Mar 2019, 13:30
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fskilnik wrote:
GMATH practice exercise (Quant Class 12)

Hi, Ian! Thanks for joining, for your kind words and for all your important comments related to the "theme"!

Let me add the following observation:

In each question, the GMAT asks the test taker to find the BEST answer (among the 5 alternative choices offered), therefore if you were given the following alternative choices:

(A) x < 3
(B) x < 2
(C) ---
(D) ---
(E) ---

You would be supposed to consider (B) the "proper" right answer (it is more "restrictive"), although (as explained by Ian) alternative choice (A) is also true. (If x < 2 then x<3, of course!)

The fact is that I did NOT offer two right answers (to have to deal with the "best" one to be chosen, something I personally dislike), therefore the official answer is (A) without any mess!

Okay... I hope this comment is a good complement to Ian´s discussion!

Now let´s go to "the official solution":

$$?\,\,\,:\,\,\,\,x\,\,{\rm{must}}\,\,{\rm{be}}\,\,\, \ldots$$

$$x + \sqrt {{x^2} - 4x + 4} \,\,\, = \,\,\,2\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\sqrt {{{\left( {x - 2} \right)}^2}} = 2 - x$$

$$\,\,\,\, \Leftrightarrow \,\,\,\,\,\left| {x - 2} \right| = - \left( {x - 2} \right)\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x - 2 \le 0\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \le 2$$

$$x \le 2\,\,\,\, \Rightarrow \,\,\,\left( A \right)\,\,{\rm{is}}\,\,{\rm{true}}\,\,\,\,\left[ {{\rm{and}}\,\,\left( B \right),\left( C \right),\left( D \right),\left( E \right)\,\,{\rm{are}}\,\,{\rm{false}}} \right]$$

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Re: If x+sqrt(x^2-4x+4)=2, then:   [#permalink] 27 Mar 2019, 13:30
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