Problem 48 - Intermediate Algebra

Question

Rationalize each denominator. See Example 4. $$ \frac{2 \sqrt{a}-3}{2 \sqrt{a}+\sqrt{b}} $$

Step-by-Step Solution

Answer

The rationalized expression is \(\frac{4a - 2 \sqrt{ab} - 6 \sqrt{a} + 3 \sqrt{b}}{4a - b}\).

Step 1: Identify the Conjugate

To rationalize the denominator \(2 \sqrt{a} + \sqrt{b}\), we first need to identify its conjugate. The conjugate of \(2 \sqrt{a} + \sqrt{b}\) is \(2 \sqrt{a} - \sqrt{b}\).

Step 2: Multiply by the Conjugate

Multiply both the numerator and the denominator of the fraction by the conjugate of the denominator: \[\frac{(2 \sqrt{a} - 3)(2 \sqrt{a} - \sqrt{b})}{(2 \sqrt{a} + \sqrt{b})(2 \sqrt{a} - \sqrt{b})}\].

Step 3: Expand the Numerator

Expand the expression in the numerator using the distributive property: \[(2 \sqrt{a} - 3)(2 \sqrt{a} - \sqrt{b}) = 4a - 2 \sqrt{ab} - 6 \sqrt{a} + 3 \sqrt{b}\].

Step 4: Simplify the Denominator

Apply the difference of squares formula to the denominator: \[(2 \sqrt{a} + \sqrt{b})(2 \sqrt{a} - \sqrt{b}) = (2 \sqrt{a})^2 - (\sqrt{b})^2 = 4a - b\].

Step 5: Write the Rationalized Expression

The fraction is now rewritten as: \[\frac{4a - 2 \sqrt{ab} - 6 \sqrt{a} + 3 \sqrt{b}}{4a - b}\]. This is the rationalized form, with no square roots in the denominator.

Key Concepts

Conjugate

The term "conjugate" refers to a special pair of expressions used in mathematics to simplify certain calculations, particularly when dealing with square roots. When you have a binomial expression with a square root, like \(2 \sqrt{a} + \sqrt{b}\), its conjugate is formed by changing the sign between the terms: \(2 \sqrt{a} - \sqrt{b}\).

This concept is particularly useful when rationalizing a denominator because multiplying a binomial by its conjugate eliminates the square root terms. The result is a difference of squares, a critical algebraic identity that simplifies the expression.

• The conjugate helps remove irrational numbers from the denominator, making the expression easier to work with and understand.
• Forming a conjugate involves changing the sign between the terms, which simplifies complex expressions in calculations involving square roots.

In essence, finding the conjugate is the first vital step in rationalizing denominators to produce simpler, more manageable expressions.

Difference of Squares

The difference of squares is an algebraic identity that describes the product of two conjugates. The formula is expressed as \(a^2 - b^2 = (a + b)(a - b)\). This method is quite useful for rationalizing denominators that contain square roots.

In the exercise example, we have the denominator \((2 \sqrt{a} + \sqrt{b})(2 \sqrt{a} - \sqrt{b})\).

• Using the difference of squares formula, this breaks down to \((2 \sqrt{a})^2 - (\sqrt{b})^2\).
• This simplifies further into \(4a - b\), effectively removing all square root terms from the denominator.

This process makes the fraction much easier to simplify and understand, as it involves straightforward subtraction and multiplication, rather than managing ineffable square roots.

Distributive Property

The distributive property is a fundamental principle of algebra that allows us to multiply a term across a sum or difference inside parentheses. It is stated as \(a(b + c) = ab + ac\).

In the context of the exercise, the distributive property is utilized to expand the numerators, \((2 \sqrt{a} - 3)(2 \sqrt{a} - \sqrt{b})\). Let's break it down:

• First, multiply \(2 \sqrt{a}\) with both terms in the second bracket: \(4a - 2 \sqrt{ab}\).
• Next, multiply \(-3\) with both terms in the second bracket: \(- 6 \sqrt{a} + 3 \sqrt{b}\).
This leads to the expanded form \(4a - 2 \sqrt{ab} - 6 \sqrt{a} + 3 \sqrt{b}\).

By applying the distributive property, we can handle each term separately and tidy up expressions for smoother calculations. This step-by-step approach is particularly helpful when simplifying mathematical expressions as it ensures all components are accurately managed.