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16 Sep 2014, 01:26
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If \(xyz \ne 0\), is \((x^{-4})*(\sqrt[3]{y})*(z^{-2}) \lt 0\)?
(1) \(\sqrt[5]{y} \gt \sqrt[4]{x^2}\)
(2) \(y^3 \gt \frac{1}{z{^4}}\)
(1) \(\sqrt[5]{y} \gt \sqrt[4]{x^2}\)
(2) \(y^3 \gt \frac{1}{z{^4}}\)
16 Sep 2014, 01:26
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Official Solution:
If \(xyz \ne 0\), is \((x^{-4})*(\sqrt[3]{y})*(z^{-2}) \lt 0\)?
\(xyz \ne 0\) means that none of the unknowns is equal to zero. Next, \((x^{-4})*(\sqrt[3]{y})*(z^{-2}) = \frac{\sqrt[3]{y}}{x^4*z^2}\), so the question becomes: is \(\frac{\sqrt[3]{y}}{x^4*z^2} \lt 0\)? Since \(x^4\) and \(z^2\) are positive numbers, the question boils down to whether \(\sqrt[3]{y} \lt 0\), which is the same as whether \(y \lt 0\) (recall that odd roots have the same sign as the base of the root, for example: \(\sqrt[3]{125} = 5\) and \(\sqrt[3]{-64} = -4\)).
(1) \(\sqrt[5]{y} \gt \sqrt[4]{x^2}\).
Since the even root of a positive number (\(x^2\) in our case) is positive, then \(\sqrt[5]{y} \gt \sqrt[4]{x^2} \gt 0\), (which is the same as saying \(y \gt 0\)). Therefore, the answer to the original question is NO. Sufficient.
(2) \(y^3 \gt \frac{1}{z{^4}}\).
Similarly, as \(\frac{1}{z^4} \gt 0\), then \(y^3 \gt \frac{1}{z^4} \gt 0\), (which is the same as saying \(y \gt 0\)). Therefore, the answer to the original question is NO. Sufficient.
Answer: D
If \(xyz \ne 0\), is \((x^{-4})*(\sqrt[3]{y})*(z^{-2}) \lt 0\)?
\(xyz \ne 0\) means that none of the unknowns is equal to zero. Next, \((x^{-4})*(\sqrt[3]{y})*(z^{-2}) = \frac{\sqrt[3]{y}}{x^4*z^2}\), so the question becomes: is \(\frac{\sqrt[3]{y}}{x^4*z^2} \lt 0\)? Since \(x^4\) and \(z^2\) are positive numbers, the question boils down to whether \(\sqrt[3]{y} \lt 0\), which is the same as whether \(y \lt 0\) (recall that odd roots have the same sign as the base of the root, for example: \(\sqrt[3]{125} = 5\) and \(\sqrt[3]{-64} = -4\)).
(1) \(\sqrt[5]{y} \gt \sqrt[4]{x^2}\).
Since the even root of a positive number (\(x^2\) in our case) is positive, then \(\sqrt[5]{y} \gt \sqrt[4]{x^2} \gt 0\), (which is the same as saying \(y \gt 0\)). Therefore, the answer to the original question is NO. Sufficient.
(2) \(y^3 \gt \frac{1}{z{^4}}\).
Similarly, as \(\frac{1}{z^4} \gt 0\), then \(y^3 \gt \frac{1}{z^4} \gt 0\), (which is the same as saying \(y \gt 0\)). Therefore, the answer to the original question is NO. Sufficient.
Answer: D
General Discussion
