Problem 62 - Precalculus: Functions and Graphs

Question

Rationalize the denominator. $$\frac{16 x^{2}-y^{2}}{2 \sqrt{x}-\sqrt{y}}$$

Step-by-Step Solution

Answer

Rationalized form is \(\frac{32 x^{2} \sqrt{x} + 16 x^{2} \sqrt{y} - 2y^{2} \sqrt{x} - y^{2} \sqrt{y}}{4x - y}\).

Step 1: Identify the expression

We need to rationalize the denominator of the expression \(\frac{16 x^{2}-y^{2}}{2 \sqrt{x}-\sqrt{y}}\). This involves removing the radicals from the denominator.

Step 2: Use the conjugate

To rationalize the denominator, multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of \(2 \sqrt{x} - \sqrt{y}\) is \(2 \sqrt{x} + \sqrt{y}\).

Step 3: Multiply the numerator and denominator

Multiply both the numerator and the denominator by the conjugate. The expression becomes:\[\frac{(16 x^{2} - y^{2})(2 \sqrt{x} + \sqrt{y})}{(2 \sqrt{x} - \sqrt{y})(2 \sqrt{x} + \sqrt{y})}\]

Step 4: Simplify the denominator

The denominator \((2 \sqrt{x} - \sqrt{y})(2 \sqrt{x} + \sqrt{y})\) is a difference of squares and simplifies to:\[(2 \sqrt{x})^2 - (\sqrt{y})^2 = 4x - y\]

Step 5: Expand the numerator

Expand the numerator \((16 x^{2} - y^{2})(2 \sqrt{x} + \sqrt{y})\). Use the distributive property (FOIL method):\[(16 x^{2})(2 \sqrt{x}) + (16 x^{2})(\sqrt{y}) - (y^{2})(2 \sqrt{x}) - (y^{2})(\sqrt{y})\]This becomes:\[32 x^{2} \sqrt{x} + 16 x^{2} \sqrt{y} - 2y^{2} \sqrt{x} - y^{2} \sqrt{y}\]

Step 6: Combine and simplify

Place the simplified numerator over the simplified denominator:\[\frac{32 x^{2} \sqrt{x} + 16 x^{2} \sqrt{y} - 2y^{2} \sqrt{x} - y^{2} \sqrt{y}}{4x - y}\]

Step 7: Final expression

The final rationalized expression is:\[\frac{32 x^{2} \sqrt{x} + 16 x^{2} \sqrt{y} - 2y^{2} \sqrt{x} - y^{2} \sqrt{y}}{4x - y}\]

Key Concepts

Conjugate

When rationalizing denominators that contain radicals, we often encounter the term "conjugate". The conjugate of an expression plays a crucial role in eliminating radicals from the denominator. If you have an expression like \(a - b\), its conjugate is \(a + b\). Essentially, you change the sign between the two terms. This technique is particularly helpful when dealing with binomials involving square roots. Conjugates transform the denominator into a simpler expression by leveraging the difference of squares approach. They allow the radicals to cancel out, leading to a cleaner, simplified result.

Difference of Squares

The difference of squares is a special algebraic identity. It is given by \((a^2 - b^2) = (a - b)(a + b)\). This identity simplifies the multiplication of a pair of conjugates. For instance, multiplying \(2\sqrt{x} - \sqrt{y}\) by its conjugate \(2\sqrt{x} + \sqrt{y}\) simplifies to \((2\sqrt{x})^2 - (\sqrt{y})^2\). This results in \(4x - y\). Using the difference of squares ensures that the resulting expression from the denominator is free of any radicals. This makes it easier to understand and further manipulate or solve.

Simplifying Radicals

Simplifying radicals involves rewriting a radical expression in its simplest form. This process includes eliminating any radicands that are perfect squares and ensuring there are no radicals in the denominator. Let's break it down:

• Identify perfect square factors: Check if the number inside the radical can be expressed as a product of a square number. For example, \(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\).
• Rationalizing: If there are radicals in the denominator, we use the conjugate as discussed earlier. Once simplified, ensure all radicals are in their simplest form.
• Combine like terms: When multiple radical terms are present, combine them as much as possible to simplify the expression.

Simplifying radicals is an essential step in making mathematical expressions easier to work with and solve. It builds a strong foundation for tackling more complex algebraic tasks.