Bunuel wrote:

If xy>0, which of the following must be positive?

\(xy > 0 \implies x > 0 \; \text{and}\; y > 0 \quad \text{or} \quad x < 0 \;\text{and}\; y< 0 \)

A. \((\sqrt{xy})^2x^3y^2\)\((\sqrt{xy})^2 > 0 \quad \forall xy > 0\)

\(x^3y^2=(xy)^2 \times x\)

If \(x < 0 \implies x^3y^2 < 0 \implies (\sqrt{xy})^2x^3y^2 < 0\)

Eliminated.

B. \((\sqrt[3]{x})(\sqrt[5]{y})x^3y^2\)\((\sqrt[3]{x})(\sqrt[5]{y})x^3y^2=(x)^{\frac{1}{3}}(y)^{\frac{1}{5}}x^3y^2=(x)^{\frac{10}{3}}(y)^{\frac{11}{5}} \)

\(=(x)^{\frac{10}{3}}(y)^{\frac{10}{5}} \times (y)^{\frac{1}{5}} \)

\(= \Big ( (x)^{\frac{1}{3}}(y)^{\frac{1}{5}} \Big ) ^{10} \times (y)^{\frac{1}{5}}\)

If \(y < 0 \implies (y)^{\frac{1}{5}} < 0 \implies \Big ( (x)^{\frac{1}{3}}(y)^{\frac{1}{5}} \Big ) ^{10} \times (y)^{\frac{1}{5}} < 0\)

Eliminated.

C. \((\sqrt{xy})x^7y^9\) \((\sqrt{xy})x^7y^9 = (xy)^{\frac{1}{2}}(xy)^7y^2 \)

We have \(\sqrt{xy} > 0\)

\((xy)^7 > 0\) because \(xy>0\)

\(y^2 > 0\) because \(y \neq 0\)

Hence \((\sqrt{xy})x^7y^9 > 0\). This is the correct answer.

D. \((\sqrt[3]{x^2y})x^2y^4\)\((\sqrt[3]{x^2y})x^2y^4=\sqrt[3]{x^2}\times\sqrt[3]{y}\times x^2y^4\)

If \(y < 0 \implies \sqrt[3]{y} < 0 \implies \sqrt[3]{x^2}\times\sqrt[3]{y}\times x^2y^4 < 0\).

Eliminated.

E. \((\sqrt{x^3y^5})x^2y^5\)\(\sqrt{x^3y^5} \times x^2y^5 = \sqrt{(xy)^3 \times y^2}\times (xy)^2y^3 = \sqrt{(xy)^3} \times \sqrt{y^2} \times (xy)^2y^3 = \sqrt{(xy)^3} \times |y| \times (xy)^2y^3\)

Note that \(\sqrt{(xy)^3} > 0\) and \((xy)^2 > 0\) and \(|y| > 0\).

If \(y < 0 \implies y^3 < 0 \). Hence this choice is wrong.

The answer is C.

_________________

Actual LSAT CR bank by Broall

How to solve quadratic equations - Factor quadratic equations

Factor table with sign: The useful tool to solve polynomial inequalities

Applying AM-GM inequality into finding extreme/absolute value

New Error Log with Timer