Problem 110 - College Algebra with Modeling and Visualization

Question

Rationalize the denominator. $$ \frac{2 \sqrt{z}}{2-\sqrt{z}} $$

Step-by-Step Solution

Answer

\(\frac{4 \sqrt{z} + 2z}{4 - z}\).

Step 1: Identify the Approach

We need to rationalize the denominator of \( \frac{2 \sqrt{z}}{2 - \sqrt{z}} \). To do this, we will multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of \(2 - \sqrt{z}\) is \(2 + \sqrt{z}\).

Step 2: Multiply by the Conjugate

Multiply both the numerator and the denominator by \(2 + \sqrt{z}\):\[\frac{2 \sqrt{z}}{2 - \sqrt{z}} \times \frac{2 + \sqrt{z}}{2 + \sqrt{z}}\]This gives:\[\frac{2 \sqrt{z} (2 + \sqrt{z})}{(2 - \sqrt{z})(2 + \sqrt{z})}\]

Step 3: Simplify the Denominator

Simplify the denominator using the difference of squares formula, \((a-b)(a+b) = a^2 - b^2\):\[(2 - \sqrt{z})(2 + \sqrt{z}) = 2^2 - (\sqrt{z})^2 = 4 - z\]

Step 4: Expand the Numerator

Distribute \(2 \sqrt{z}\) in the numerator:\[2 \sqrt{z} (2 + \sqrt{z}) = 4 \sqrt{z} + 2z\]

Step 5: Write the Final Expression

Now that both the numerator and the denominator are expanded, write out the expression:\[\frac{4 \sqrt{z} + 2z}{4-z}\]

Step 6: Confirm the Rationalization

Confirm that the denominator \(4 - z\) has no radicals. Thus, the denominator is rationalized. The final expression is\[\frac{4 \sqrt{z} + 2z}{4 - z}\]

Key Concepts

Conjugates

When working with expressions that involve radicals, especially in denominators, rationalizing them is often necessary. An invaluable tool here is the concept of conjugates. A conjugate involves two binomials that are identical except for the sign between them changing; for example, if one binomial is \( a - b \), the conjugate will be \( a + b \).

The usefulness of conjugates comes from their property that, when multiplied together, they eliminate radicals thanks to the "difference of squares" principle (which we'll cover later). This property makes them essential in simplifying expressions to have rational denominators.

In the given exercise, the conjugate of the denominator \(2 - \sqrt{z}\) is \(2 + \sqrt{z}\). By multiplying both the numerator and the denominator by the conjugate, we end up with a rational denominator. This is because

• Conjugates result in a simple subtraction of squares when multiplied \((a^2 - b^2)\).
• They help transform expressions into a form where radicals in the denominator are eliminated.

Difference of Squares

The difference of squares is a helpful algebraic identity used when dealing with conjugates. The formula states:\[(a - b)(a + b) = a^2 - b^2\] This means when you multiply two conjugates together, the result is always a difference of squares.

In the exercise example, the denominator \((2 - \sqrt{z})(2 + \sqrt{z})\) simplifies according to this rule. Here \( a = 2 \) and \( b = \sqrt{z} \). Hence,

• \((2 - \sqrt{z})(2 + \sqrt{z}) = 2^2 - (\sqrt{z})^2\)
• This calculates to \(4 - z\).

By translating the original problem into this expression, we smoothly handled the radicals in the denominator. Resultantly, the difference of squares is a crucial step in rationalizing denominators as it quickly deals with the radical part.

Radicals

Radicals, especially square roots, often make expressions unwieldy, notably when they appear in the denominator. This is why rationalizing a denominator becomes important. To "rationalize" means to eliminate any radicals, transforming them into rational numbers.

To address this in expressions, one typically employs techniques like multiplying by the conjugate, as seen in the exercise. Radicals follow specific mathematical rules, such as:

• \(\sqrt{x} \times \sqrt{x} = x\), which allows squaring the radical terms away.
• They are treated in the hierarchy of operations, similar to exponents.

In summary, rationalizing the denominator where radicals are present gives us expressions that are easier to manage for further calculations or evaluations, avoiding complications that radicals introduce in terms of both calculation and simplification.