Carcass wrote:
If \(a, b, c, d,\) and z are non-zero integers such that \((az) ^{79} * (bz) ^{71} * (cz) ^8 * (dz) ^2 > 0\), which of the following must be true?
(A) \(z > 0\)
(B) \(ab > 0 \)
(C) \(abcd > 0 \)
(D) \(b^2 – a^2 > 0 \)
(E) \(az + bz> 0 \)
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITIONGRE - Math Book Key property: Any non-zero number raised to an EVEN power will evaluate to be a POSITIVE number. Given: \((az) ^{79} \times (bz) ^{71} \times (cz) ^8 \times (dz) ^2 > 0\)
Rewrite as: \(a ^{79} \times z^{79} \times b^{71} \times z^{71} \times c^8 \times z^8 \times d^2 \times z^2 > 0\)
Combine all of the \(z\) terms: \(a ^{79} \times b^{71} \times c^8 \times d^2 \times z^{160} > 0\)
Since we can be certain that \(c^8\) is positive, we can safely divide both sides of the inequality by \(c^8\) to get: \(a ^{79} \times b^{71} \times d^2 \times z^{160} > 0\)
For the same reason, we can safely divide both sides of the inequality by \(d^2\) to get: \(a ^{79} \times b^{71} \times z^{160} > 0\)
And, for the same reason, we can safely divide both sides of the inequality by \(z^{160}\) to get: \(a ^{79} \times b^{71} > 0\)
Since we can be certain that \(a^{78}\) is positive, we can safely divide both sides of the inequality by \(a^{78}\) to get: \(a ^{1} \times b^{71} > 0\)
Since we can be certain that \(b^{70}\) is positive, we can safely divide both sides of the inequality by \(b^{70}\) to get: \(a ^{1} \times b^{1} > 0\)
In other words, \(ab > 0\)
Answer: B
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Brent Hanneson - founder of Greenlight Test Prep