Problem 102 - College Algebra with Modeling and Visualization

Question

Rationalize the denominator. $$ \sqrt{\frac{5 b}{72}} $$

Step-by-Step Solution

Answer

The rationalized form is \( \frac{\sqrt{10b}}{12} \).

Step 1: Express the Square Root Properly

Rewrite the expression with the square root applied separately to the numerator and the denominator: \[ \sqrt{\frac{5b}{72}} = \frac{\sqrt{5b}}{\sqrt{72}} \]

Step 2: Simplify the Denominator

Recognize that \( \sqrt{72} \) can be simplified. Since \( 72 = 36 \times 2 \) and \( \sqrt{36} = 6 \), we have: \[ \sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2} \] Now substitute back: \[ \frac{\sqrt{5b}}{\sqrt{72}} = \frac{\sqrt{5b}}{6\sqrt{2}} \]

Step 3: Rationalize the Denominator

To rationalize the denominator, multiply the numerator and the denominator by \( \sqrt{2} \): \[ \frac{\sqrt{5b}}{6\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{5b} \cdot \sqrt{2}}{6 \cdot 2} = \frac{\sqrt{10b}}{12} \]

Step 4: Simplified Rationalized Form

After rationalization, the final expression is: \[ \frac{\sqrt{10b}}{12} \] The denominator is now rational.

Key Concepts

Square Roots

Square roots are fundamental in mathematics, representing one of two equal factors of a number. When you take the square root of a number, you're identifying what number, when multiplied by itself, gives you the initial number. For example, the square root of 25 is 5 because \( 5 \times 5 = 25 \).In the context of the exercise where we want to rationalize the denominator of \( \sqrt{\frac{5b}{72}} \), we start by expressing the square root separately for both the numerator and the denominator: \[ \sqrt{\frac{5b}{72}} = \frac{\sqrt{5b}}{\sqrt{72}} \]. It's crucial to do so because it allows us to simplify the expression more easily. The square root of a fraction is equivalent to the square root of the numerator divided by the square root of the denominator.

Simplifying Radicals

Simplifying radicals involves expressing a square root in its simplest form. This process makes calculations more manageable and expressions easier to understand. Let's explore how it works using our example.First, notice that the denominator, \( \sqrt{72} \), is not in its simplest form. We simplify it by factoring 72 into its prime factors:

• \( 72 = 2 \times 36 = 2 \times 6 \times 6 \)
• \( \sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} \)
• Since \( \sqrt{36} = 6 \), this simplifies to \( 6\sqrt{2} \).

Substitute \( 6\sqrt{2} \) back into the expression: \( \frac{\sqrt{5b}}{\sqrt{72}} = \frac{\sqrt{5b}}{6\sqrt{2}} \). This provides a clearer path to rationalization, maintaining the mathematical integrity of the expression.

Rational Numbers

A rational number is any number that can be expressed as a quotient or fraction of two integers, where the denominator is not zero. Rationalizing the denominator means converting a fraction so that the denominator becomes a rational number.In our iterative process to ensure the expression \( \frac{\sqrt{5b}}{6\sqrt{2}} \) has a rational denominator, we multiply both the numerator and the denominator by the radical present on the denominator, in this case \( \sqrt{2} \): \[ \frac{\sqrt{5b} \times \sqrt{2}}{6\sqrt{2} \times \sqrt{2}} = \frac{\sqrt{10b}}{12} \]. By doing this, the denominator becomes \( 12 \), a rational number, because \( \sqrt{2} \times \sqrt{2} = 2 \) and multiplying that with the 6 gives us a non-radical \( 12 \). This rational form is easier to interpret and use in further calculations.