Problem 13 - College Algebra with Corequisite Support

Question

For the following exercises, simplify each expression. \(\sqrt{\frac{81}{5}}\)

Step-by-Step Solution

Answer

The simplified expression is \( \frac{9\sqrt{5}}{5} \).

Step 1: Identify the expression inside the square root

The expression given is \( \sqrt{\frac{81}{5}} \). We need to simplify this square root expression.

Step 2: Separate the square root

Use the property of square roots that \( \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} \). So, \( \sqrt{\frac{81}{5}} = \frac{\sqrt{81}}{\sqrt{5}} \).

Step 3: Simplify the square root of the numerator

Calculate \( \sqrt{81} \). Since 81 is a perfect square (\( 9 \times 9 = 81 \)), \( \sqrt{81} = 9 \).

Step 4: Simplify the square root of the denominator

The denominator is \( \sqrt{5} \). Since 5 is not a perfect square, \( \sqrt{5} \) remains as it is.

Step 5: Form the simplified expression

Now substitute back the simplified components to get \( \frac{9}{\sqrt{5}} \).

Step 6: Rationalize the denominator

Multiply both numerator and denominator by \( \sqrt{5} \) to rationalize the denominator. Thus, \( \frac{9}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{9\sqrt{5}}{5} \).

Step 7: Write the final simplified expression

The simplified expression of \( \sqrt{\frac{81}{5}} \) is \( \frac{9\sqrt{5}}{5} \).

Key Concepts

Rationalizing Denominators

Rationalizing denominators is a crucial step in simplifying square root expressions. This process eliminates any square roots present in the denominator of a fraction. By doing so, we make the expression easier to interpret and work with, especially for further calculations or comparisons.

In the context of our exercise, the expression \( \frac{9}{\sqrt{5}} \) necessitates rationalization. Since there is a square root in the denominator, we multiply both the numerator and denominator by \( \sqrt{5} \). This technique leverages the identity \( (\sqrt{5})^2 = 5 \), which transforms the denominator into a rational number.

• Multiply the top and bottom by \( \sqrt{5} \).
• Result: \( \frac{9\sqrt{5}}{5} \).

This method maintains the value of the expression but converts it into a more acceptable and standardized form. Rationalizing makes subsequent mathematical operations more straightforward.

Properties of Square Roots

Understanding the properties of square roots is fundamental to simplifying expressions effectively. These properties allow us to manipulate and break down complex square root expressions into simpler forms.

One key property applied in our exercise is \( \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} \). This property helps us deal with fractions easily by separating the square root of a fraction into the division of two separate square roots.

Here's how this property was used:

• We started with \( \sqrt{\frac{81}{5}} \) and rewrote it as \( \frac{\sqrt{81}}{\sqrt{5}} \).
• The square root of the perfect square (\( \sqrt{81} = 9 \)) was straightforward, reducing complexity.
• The square root of non-perfect squares like \( \sqrt{5} \) remain unchanged for simplicity.

Mastering these properties leads to quick and accurate expressions, minimizing potential errors.

Perfect Squares

Perfect squares are numbers whose square roots are whole numbers. Recognizing perfect squares is critical in simplifying square root expressions since they simplify to a whole number, making expressions cleaner and more concise.

In our exercise, \( 81 \) is identified as a perfect square because it equals \( 9 \times 9 \). Therefore, \( \sqrt{81} \) simplifies to \( 9 \). This significantly simplifies our calculations and helps us focus on more challenging parts of the expression.

Other examples of perfect squares include 4, 16, 25, 64, etc. Memorizing these can greatly speed up the simplification process:

• \( 4 = 2 \times 2 \rightarrow \sqrt{4} = 2 \)
• \( 16 = 4 \times 4 \rightarrow \sqrt{16} = 4 \)
• \( 25 = 5 \times 5 \rightarrow \sqrt{25} = 5 \)
• \( 64 = 8 \times 8 \rightarrow \sqrt{64} = 8 \)

Recognizing and using perfect squares simplifies expressions considerably, allowing for quicker problem-solving.