Question

5 which expression is equivalent to \\(sqrt4{xy} cdot sqrt4{2x}\\) where \\(x\\) and \\(y\\) are positive? hint \\(\bigcirc\\) a. \\(sqrt4{2x^3 y}\\) \\(\bigcirc\\) b. \\(sqrt4{2x^2 y^4}\\) \\(\bigcirc\\) c. \\(sqrt4{10x^2 y}\\) \\(\bigcirc\\) d. \\(sqrt4{16x^2 y^4}\\)

Explanation:

Step1: Convert radicals to exponents

Recall that \(\sqrt[n]{a}=a^{\frac{1}{n}}\). So, \(\sqrt[4]{xy}=(xy)^{\frac{1}{4}}\) and \(\sqrt[4]{2x}=(2x)^{\frac{1}{4}}\).

Step2: Multiply the two expressions

When multiplying two exponents with the same base (here, the base is the fourth root, so we can multiply the radicands), so \((xy)^{\frac{1}{4}}\cdot(2x)^{\frac{1}{4}}=(xy\cdot2x)^{\frac{1}{4}}\).

Step3: Simplify the radicand

Multiply the terms inside the radicand: \(xy\cdot2x = 2x^{1 + 1}y=2x^{2}y\). Then, convert back to radical form: \((2x^{2}y)^{\frac{1}{4}}=\sqrt[4]{2x^{2}y}\). Wait, no, wait, let's check again. Wait, maybe I made a mistake in the exponent. Wait, the original problem: \(\sqrt[4]{xy}\cdot\sqrt[4]{2x}\). Let's do it by multiplying the radicands directly since \(\sqrt[n]{a}\cdot\sqrt[n]{b}=\sqrt[n]{ab}\) (property of radicals: product of nth roots is nth root of product). So \(\sqrt[4]{xy}\cdot\sqrt[4]{2x}=\sqrt[4]{(xy)\cdot(2x)}\). Then multiply the terms: \(x\cdot x = x^{2}\), \(y\) remains, and 2 remains. So \((xy)(2x)=2x^{2}y\). So \(\sqrt[4]{2x^{2}y}\). Wait, but let's check the options. Option A is \(\sqrt[4]{2x^{2}y}\), option B is \(\sqrt[4]{2x^{2}y^{2}}\), option C is \(\sqrt[4]{10x^{2}y}\), option D is \(\sqrt[4]{16x^{2}y^{2}}\). Wait, maybe I misread the original problem. Wait, maybe the first radical is \(\sqrt[4]{x y^3}\)? Wait, no, the user's image: the problem is \(\sqrt[4]{xy}\cdot\sqrt[4]{2x}\)? Wait, no, maybe the first term is \(\sqrt[4]{x y^3}\)? Wait, no, let's check the options. Option B is \(\sqrt[4]{2x^{2}y^{2}}\), option A is \(\sqrt[4]{2x^{2}y}\). Wait, maybe I made a mistake. Wait, let's re-express the radicals correctly. Let's use exponent rules. \(\sqrt[4]{xy}=x^{\frac{1}{4}}y^{\frac{1}{4}}\), \(\sqrt[4]{2x}=2^{\frac{1}{4}}x^{\frac{1}{4}}\). Multiply them: \(x^{\frac{1}{4}}y^{\frac{1}{4}}\cdot2^{\frac{1}{4}}x^{\frac{1}{4}}=2^{\frac{1}{4}}x^{\frac{1}{4}+\frac{1}{4}}y^{\frac{1}{4}}=2^{\frac{1}{4}}x^{\frac{2}{4}}y^{\frac{1}{4}}=2^{\frac{1}{4}}x^{\frac{1}{2}}y^{\frac{1}{4}}\). Now, let's convert each option to exponents:

Option A: \(\sqrt[4]{2x^{2}y}=2^{\frac{1}{4}}x^{\frac{2}{4}}y^{\frac{1}{4}}=2^{\frac{1}{4}}x^{\frac{1}{2}}y^{\frac{1}{4}}\), which matches.

Option B: \(\sqrt[4]{2x^{2}y^{2}}=2^{\frac{1}{4}}x^{\frac{2}{4}}y^{\frac{2}{4}}=2^{\frac{1}{4}}x^{\frac{1}{2}}y^{\frac{1}{2}}\), which has \(y^{\frac{1}{2}}\) instead of \(y^{\frac{1}{4}}\), so no.

Option C: \(\sqrt[4]{10x^{2}y}=10^{\frac{1}{4}}x^{\frac{2}{4}}y^{\frac{1}{4}}\), coefficient is 10, not 2, so no.

Option D: \(\sqrt[4]{16x^{2}y^{2}}=16^{\frac{1}{4}}x^{\frac{2}{4}}y^{\frac{2}{4}}=2x^{\frac{1}{2}}y^{\frac{1}{2}}\), coefficient 2, but \(y\) exponent is \(\frac{1}{2}\), so no.

So the correct answer is A.

Answer:

A. \(\sqrt[4]{2x^{2}y}\)