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How do you know which route to pick?

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How do you know which route to pick? [#permalink]  25 Dec 2010, 11:27
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Question below is simple. I know the answer and I know how to solve it but if you begin with squaring both sides, there comes a point in the problem when: x + 3$$\sqrt{x}$$+3 = 0. From this point on, I can set up the equation two ways to solve for x^2:

1) x = -3$$\sqrt{x}$$ - 3
or
2) -3$$\sqrt{x}$$ = x+3

Route (2) gives me the answer while route (1) does not. So my questions are, how am I suppose to know which route to take during the exam? Is this an art more than the science? Do 700 students automatically know which route to take? Am I missing an important trick/concept here?

53. If $$\sqrt{3-x}$$ = $$\sqrt{x}$$ + 3, then x^2 =
(A) 1
(B) 3
(C) 2 - 3x
(D) 3x - 1
(E) 3x - 9
[Reveal] Spoiler: OA
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Re: How do you know which route to pick? [#permalink]  25 Dec 2010, 15:36
Expert's post
Something is wrong with the equation:
$$\sqrt{3-x} = \sqrt{x} + 3$$

the right side is always greater or equal 3 and x must be non-negative. At the same time the left side is less or equal $$\sqrt{3}$$ for non-negative x.
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Re: How do you know which route to pick? [#permalink]  26 Dec 2010, 11:22
What's the source of this question?
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Re: How do you know which route to pick? [#permalink]  26 Dec 2010, 11:38
GMATHacks - 10 sample questions
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Re: How do you know which route to pick? [#permalink]  26 Dec 2010, 11:50
Walker - At what point did you decide to test the question itself? Was it when you saw that setting up one equation in two different ways, only one gets you to the answer? I need to learn how to think outside of box. Thanks
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Re: How do you know which route to pick? [#permalink]  26 Dec 2010, 12:14
Expert's post
Yeah, I found it.
So, there is no typo in your post but as I said there is no x that satisfies the equation. (GMAT doesn't deal with complex numbers).

When I decided to test the question? Hm... I often "play" with numbers before solving a problem. Sometimes it helps to find a shortcut.

Suppose we have the equation: a = b. Can we say that a^2 = b^2? Yeah, it's correct. Now, let's take a look at an example:

$$-\sqrt{x} = 2$$
$$(-\sqrt{x})^2 = 2^2$$
$$x = 4$$

Is it right? No. Because $$-\sqrt{4} = -2 \neq 2$$. So, a solution of a^2=b^2 isn't necessarily a solution of a = b and it's important to check it out. So, x=4 is a correct solution of $$(-\sqrt{x})^2 = 2^2$$ but it's not a solution of $$-\sqrt{x} = 2$$.

Hope it helps
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Re: How do you know which route to pick? [#permalink]  26 Dec 2010, 13:52
Wait....you mentioned that -$$\sqrt{4}$$= -2 but since you can interpret -$$\sqrt{4}$$ as (-1) * $$\sqrt{4}$$ = (-1) * [either(-2) or 2] and therefore -$$\sqrt{4}$$ = actually either 2 or (-2). No?
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Re: How do you know which route to pick? [#permalink]  26 Dec 2010, 14:08
Expert's post
http://en.wikipedia.org/wiki/Square_root

"Every positive number x has two square roots. One of them is $$\sqrt{x}$$, which is positive, and the other $$-\sqrt{x}$$, which is negative."
"Every non-negative real number x has a unique non-negative square root, called the principal square root, denoted by a radical sign as $$\sqrt{x}$$.

$$\sqrt{x}$$ is the principal square root and non-negative. So $$-\sqrt{4}$$ means only -2 because a radical sign means only non-negative square root.
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Re: How do you know which route to pick?   [#permalink] 26 Dec 2010, 14:08
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