Question

cphs : advanced algebra: concepts and connections - block (27.0831030)
dividing radicals
rationalizing the denominator
which expression should you multiply the numerator and denominator of \\(\frac{\sqrt3{3}}{\sqrt3{2x}}\\) by to rationalize the denominator?
\\(\sqrt3{2x}\\)
\\(\sqrt3{x^2}\\)
\\(\sqrt3{4x}\\)
\\(\sqrt3{4x^2}\\)

Explanation:

Step1: Analyze the denominator

The denominator is $\sqrt[3]{2x}$. For a cube root, we need to make the radicand a perfect cube. The radicand here is $2x$. To make it a perfect cube, we need to find what to multiply $2x$ by to get a perfect cube. Let's factor the current radicand: $2x = 2 \times x$. We need the exponents of all prime factors (and variables) to be multiples of 3. So for 2, we need two more 2s (since $2^1$ needs to become $2^3$), and for $x$, we need two more $x$s (since $x^1$ needs to become $x^3$). Wait, no, actually, let's look at the options. Wait, the fraction is $\frac{\sqrt[3]{3}}{\sqrt[3]{2x}}$. To rationalize the denominator (cube root), we need to multiply numerator and denominator by a radical that will make the denominator's radicand a perfect cube. The denominator's radicand is $2x$. Let's denote the denominator as $\sqrt[3]{2x}=\sqrt[3]{2 \times x}$. We need to multiply by $\sqrt[3]{4x^2}$ because $2x \times 4x^2 = 8x^3$, which is a perfect cube ($8x^3=(2x)^3$). Let's check: $\sqrt[3]{2x} \times \sqrt[3]{4x^2}=\sqrt[3]{2x \times 4x^2}=\sqrt[3]{8x^3}=2x$, which is rational (no radical in the denominator). Let's check the other options:

• Option with $\sqrt[3]{2x}$: Multiplying by $\sqrt[3]{2x}$ would give $\sqrt[3]{2x \times 2x}=\sqrt[3]{4x^2}$, not a perfect cube.
• Option with $\sqrt[3]{x^2}$: $\sqrt[3]{2x} \times \sqrt[3]{x^2}=\sqrt[3]{2x \times x^2}=\sqrt[3]{2x^3}=x\sqrt[3]{2}$, still has a radical.
• Option with $\sqrt[3]{4x}$: $\sqrt[3]{2x} \times \sqrt[3]{4x}=\sqrt[3]{8x^2}=\sqrt[3]{8 \times x^2}=2\sqrt[3]{x^2}$, still has a radical.
• Option with $\sqrt[3]{4x^2}$: As above, $\sqrt[3]{2x} \times \sqrt[3]{4x^2}=\sqrt[3]{8x^3}=2x$, which is rational.

Step2: Confirm the multiplication

So we need to multiply numerator and denominator by $\sqrt[3]{4x^2}$ to rationalize the denominator.

Answer:

The expression is $\sqrt[3]{4x^2}$ (the first option on the left, the rectangle with $\sqrt[3]{4x^2}$).