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25 Dec 2010, 12:27
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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
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
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25 Dec 2010, 16:36
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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.
\(\sqrt{3-x} = \sqrt{x} + 3\)
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.
