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19 May 2020, 03:46
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
49% (02:28) correct
51%
(02:25)
wrong
based on 150
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\(x, a, b, c\), and \(d\) are rational numbers satisfying \(\frac{1}{x + \sqrt{2} + \sqrt{3}x + \sqrt{6}} = a + \sqrt{2}b + \sqrt{3}c + \sqrt{6}d. \) What is the expression of \(a\) in terms of \(x\)?
A. \(0\)
B. \(\frac{-x}{2} \)
C. \(\frac{x}{2 } \)
D. \(\frac{-x}{2(x^2 - 2)} \)
E. \(\frac{x}{2(x^2 - 2)}\)
A. \(0\)
B. \(\frac{-x}{2} \)
C. \(\frac{x}{2 } \)
D. \(\frac{-x}{2(x^2 - 2)} \)
E. \(\frac{x}{2(x^2 - 2)}\)
19 May 2020, 04:54
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\(x + \sqrt{2} + \sqrt{3}x + \sqrt{6} \)
\(= x + \sqrt{2} + \sqrt{3} (x + \sqrt{2}\)
\(= (x + \sqrt{2} ) (1+ \sqrt{3})\)
Now,
\((x + \sqrt{2} ) =\) \(\\
(x + \sqrt{2} )*\frac{(x- \sqrt{2} ) }{ (x - \sqrt{2} )}\)
\(= \frac{x^2-2}{(x - \sqrt{2} )}\)
Also,
\((1 + \sqrt{3} ) =\)
\((1 + \sqrt{3} )*\frac{( \sqrt{3}-1 ) }{ ( \sqrt{3}-1 )}\)
\(= \frac{2}{( \sqrt{3}-1 )}\)
\(\frac{1}{x + \sqrt{2} + \sqrt{3}x + \sqrt{6}} \)
\( = \frac{(x - \sqrt{2} ) }{ x^2-2) }* \frac{( \sqrt{3}-1 )}{2}\)
\(= \frac{1}{ 2*( x^2-2)}[ (x - \sqrt{2} ) ) * ( \sqrt{3}-1 )]\)
a is free of any radical; hence multiply radial free terms to get a. [that is x*(-1)]
\( a = \frac{-x}{ 2*( x^2-2)}\)
\(= x + \sqrt{2} + \sqrt{3} (x + \sqrt{2}\)
\(= (x + \sqrt{2} ) (1+ \sqrt{3})\)
Now,
\((x + \sqrt{2} ) =\) \(\\
(x + \sqrt{2} )*\frac{(x- \sqrt{2} ) }{ (x - \sqrt{2} )}\)
\(= \frac{x^2-2}{(x - \sqrt{2} )}\)
Also,
\((1 + \sqrt{3} ) =\)
\((1 + \sqrt{3} )*\frac{( \sqrt{3}-1 ) }{ ( \sqrt{3}-1 )}\)
\(= \frac{2}{( \sqrt{3}-1 )}\)
\(\frac{1}{x + \sqrt{2} + \sqrt{3}x + \sqrt{6}} \)
\( = \frac{(x - \sqrt{2} ) }{ x^2-2) }* \frac{( \sqrt{3}-1 )}{2}\)
\(= \frac{1}{ 2*( x^2-2)}[ (x - \sqrt{2} ) ) * ( \sqrt{3}-1 )]\)
a is free of any radical; hence multiply radial free terms to get a. [that is x*(-1)]
\( a = \frac{-x}{ 2*( x^2-2)}\)
MathRevolution
21 May 2020, 19:06
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=>
\(\frac{1}{x+\sqrt{2}+\sqrt{3}x+\sqrt{6}}\)
\(=\frac{1}{x + \sqrt{2} + \sqrt{3}(x + \sqrt{2})}\)
(taking out a common factor of \(\sqrt{3}\) from the last two
terms in the denominator)
=\(\frac{1}{(x + \sqrt{2})(1 + \sqrt{3})}\) (taking out a common factor of (\(x + \sqrt{2}\)))
\(=\frac{(\sqrt{3} - 1)(x - \sqrt{2})}{(x + \sqrt{2})(x - \sqrt{2})(\sqrt{3}+1)(\sqrt{3}-1))}\) (multiplying the numerator and denominator by the conjugates \((x - \sqrt{2})(\sqrt{3}-1)\)
\(=\frac{ \sqrt{3}x - \sqrt{6} - x + \sqrt{2}}{(x^2- \sqrt{2}x + \sqrt{2}x - 2)(3 - \sqrt{3} + \sqrt{3} - 1)}\) (multiplying)
\(= \frac{-x + \sqrt{2} + \sqrt{3}x - \sqrt{6}}{(x^2- 2)(2) }\)(rearranging the numerator and adding like terms in the denominator)
\(= \frac{-x + \sqrt{2} + \sqrt{3}x - \sqrt{6}}{2(x^2 - 2)}\) (rearranging the denominator)
Thus, we have \(a = \frac{-x}{2(x^2-2)} = \frac{-x}{2(x^2-2)}\)
Therefore, the answer is D.
Answer: D
\(\frac{1}{x+\sqrt{2}+\sqrt{3}x+\sqrt{6}}\)
\(=\frac{1}{x + \sqrt{2} + \sqrt{3}(x + \sqrt{2})}\)
(taking out a common factor of \(\sqrt{3}\) from the last two
terms in the denominator)
=\(\frac{1}{(x + \sqrt{2})(1 + \sqrt{3})}\) (taking out a common factor of (\(x + \sqrt{2}\)))
\(=\frac{(\sqrt{3} - 1)(x - \sqrt{2})}{(x + \sqrt{2})(x - \sqrt{2})(\sqrt{3}+1)(\sqrt{3}-1))}\) (multiplying the numerator and denominator by the conjugates \((x - \sqrt{2})(\sqrt{3}-1)\)
\(=\frac{ \sqrt{3}x - \sqrt{6} - x + \sqrt{2}}{(x^2- \sqrt{2}x + \sqrt{2}x - 2)(3 - \sqrt{3} + \sqrt{3} - 1)}\) (multiplying)
\(= \frac{-x + \sqrt{2} + \sqrt{3}x - \sqrt{6}}{(x^2- 2)(2) }\)(rearranging the numerator and adding like terms in the denominator)
\(= \frac{-x + \sqrt{2} + \sqrt{3}x - \sqrt{6}}{2(x^2 - 2)}\) (rearranging the denominator)
Thus, we have \(a = \frac{-x}{2(x^2-2)} = \frac{-x}{2(x^2-2)}\)
Therefore, the answer is D.
Answer: D
