Problem 70 - Algebra 2

Question

How can you write \(\sqrt[3]{\frac{5}{2 x y}}\) with a rationalized denominator? \(\mathrm{F} \cdot \frac{\sqrt[3]{5}}{2 \times 4}\) G. \(\frac{\sqrt[3]{20}}{2 x y}\) H. \(\frac{\sqrt[3]{20 x^{2} y^{2}}}{2 x y}\) J. \(\frac{\sqrt[3]{4 x^{2} y^{2}}}{2 x y}\)

Step-by-Step Solution

Answer

The rationalized form of \(\sqrt[3]{\frac{5}{2xy}}\) is \(\frac{\sqrt[3]{20x^{2}y^{2}}}{2xy}\). Therefore, the solution is H.

Step 1: Determine the Cube Squares

The first step is to square the denominator so that when it is multiplied by itself, the cube root can be eliminated. The denominator is \(2xy\), hence squaring it results in \(4x^{2}y^{2}\).

Step 2: Multiply the numerator and denominator by the square of the denominator

Multiply the numerator and denominator by the cube square of the denominator, \(4x^{2}y^{2}\), to eliminate the cube root from the denominator. We get \(\frac{\sqrt[3]{5} \times \sqrt[3]{(4x^{2}y^{2})^2}}{2xy\times 4x^{2}y^{2}}\). Simplifying the expression gives \(\frac{\sqrt[3]{20x^{2}y^{2}}}{8x^{3}y^{3}}\).

Step 3: Cancel terms

Reducing the expression, some terms can be cancelled out, and it simplifies to \(\frac{\sqrt[3]{20x^{2}y^{2}}}{2xy}\). This is the final answer, which matches option H.

Key Concepts

Cube Roots

Cube roots can be a bit tricky initially, but they're just another way to "unfold" a number that has been cubed. To break it down:

• The cube root of a number \(a\) is a value \(b\) such that \(b^3 = a\).
• In our example, \(\sqrt[3]{5}\) and \(\sqrt[3]{20x^{2}y^{2}}\) are cube roots that need rationalizing.

Essentially, you're figuring out what number, when multiplied by itself three times, results in the original number. Also, the cube root applies to both numerator and denominator here, making it crucial to simplify properly.

Rationalizing Denominators

Rationalizing the denominator is a process used to make sure the expression has a rational number in the denominator. This is important in mathematical expressions to avoid dealing with irrational numbers at the bottom.

• Rationalizing typically involves multiplying the numerator and the denominator by a value that will cancel out any roots or irrational parts in the denominator.
• In this specific example, the goal was to remove the cube root from \(2xy\) in the denominator to achieve a more simplified, rational expression.

Using the concept of multiplying by the "cube square" of the denominator - \(4x^{2}y^{2}\) - helps in making the denominator rational, leading to an expression that is easier to manage.

Simplifying Expressions

Simplifying expressions is a crucial step in any algebraic procedure. It allows you to break down a complex equation into its simplest form.

• This involves combining like terms and reducing fractions to their lowest terms possible.
• In our exercise, simplifying involves both simplifying the cube root expressions and canceling out common terms.

The expression \(\sqrt[3]{5} \cdot \sqrt[3]{(4x^{2}y^{2})^2}\) gets simplified to \(\sqrt[3]{20x^{2}y^{2}}\). Similarly, simplifying the entire fraction is essential to reach the final answer, showing the neatest and most efficient form of our equation.

Numerator and Denominator Manipulation

Manipulating the numerator and denominator is an operation that is regularly performed to simplify expressions further. This involves adjusting either side to either eliminate roots or make calculations more straightforward.

• In this exercise, numerator and denominator manipulation involved multiplying both parts of the fraction by \(4x^{2}y^{2}\). This serves to cancel out the cube root in the denominator.
• After multiplication, \(\frac{\sqrt[3]{20x^{2}y^{2}}}{8x^{3}y^{3}}\) was achieved. Then, reducing the terms gave us the simplified form \(\frac{\sqrt[3]{20x^{2}y^{2}}}{2xy}\).

This manipulation ensures that the fraction is not only easier to understand but also matches the simplified forms typically required in mathematics.