Problem 49 - College Algebra Essentials

Question

In Exercises \(45-54,\) rationalize the denominator. $$\frac{13}{3+\sqrt{11}}$$

Step-by-Step Solution

Answer

The simplified form with rationalized denominator is \(-\frac{39}{2} + \frac{13\sqrt{11}}{2}\).

Step 1: Identify the conjugate

First identify the conjugate of the denominator. The conjugate of \(3 + \sqrt{11}\) is \(3 - \sqrt{11}\).

Step 2: Multiply by the conjugate

To remove the square root from the denominator, multiply both the numerator and the denominator by the conjugate. So, multply \(\frac{13}{3 + \sqrt{11}}\) by \(\frac{3 - \sqrt{11}}{3 - \sqrt{11}}\). This gives \(\frac{13 * (3 - \sqrt{11})}{(3 + \sqrt{11}) * (3 - \sqrt{11})}\).

Step 3: Simplify the expression

Now simplify the above expression. In the numerator distribute 13. In the denominator, use the difference of squares formula which is \((a - b) * (a + b) = a^2 - b^2\). Applying these facts gives \(\frac{39 - 13*\sqrt{11}}{9 - 11}\). Further simplifying gives \(\frac{39 - 13\sqrt{11}}{-2}\).

Step 4: Final simplification

Divide each term in the numerator by -2 to get \(-\frac{39}{2} + \frac{13\sqrt{11}}{2}\).

Key Concepts

Conjugate

Rationalizing a denominator with a square root involves using the conjugate. The conjugate is a neat algebraic technique that can help simplify expressions. For any expression of the form \(a + b \sqrt{c}\), the conjugate is \(a - b \sqrt{c}\). The beauty of conjugates is that multiplying a number by its conjugate results in a difference of squares.

This property makes it incredibly useful for removing square roots from a denominator. By multiplying the numerator and the denominator of a fraction by the conjugate of the denominator, you can transform the original expression into a more manageable form. This action is essential in various algebraic operations, especially when you need to rationalize a denominator containing a square root.

Difference of Squares

The difference of squares formula is a powerful tool in algebra that simplifies expressions quickly and efficiently. It is based on the identity: \[ (a + b)(a - b) = a^2 - b^2 \]This formula is especially useful in the context of rationalizing denominators.

In the case of our example, multiplying \((3 + \sqrt{11})\) by its conjugate \((3 - \sqrt{11})\) results in the difference of squares. So, the denominator simplifies to:- \(3^2 - (\sqrt{11})^2 = 9 - 11\)- This simplifies further to \(-2\).

When dealing with square roots, the difference of squares formula helps eliminate the root, making expressions more straightforward to work with.

Simplification

Simplification is the process of making an expression easier to read or solve. It's often the final goal in algebraic manipulation. For rationalizing denominators, simplification involves reducing complex fractions into simpler ones.

Once you've multiplied the numerator and denominator by the conjugate, and applied the difference of squares, you're left with a fraction that's likely easier to handle. For instance:- After multiplying, our numerator is \(39 - 13 \sqrt{11}\).- The denominator is \(-2\), simplifying the fraction to:\[ \frac{39 - 13 \sqrt{11}}{-2} \]

This can be further simplified by distributing the division across terms in the numerator, where dividing each term by \(-2\) gives us the final simplified form of the expression.

Algebraic Fractions

Algebraic fractions are fractions that contain algebraic expressions in the numerator, the denominator, or both. Understanding how to manage them is crucial in algebra, particularly when dealing with rationalizing denominators.

To work effectively with algebraic fractions, you need to be comfortable with operations like multiplying by conjugates or applying the difference of squares. These techniques help simplify expressions significantly. In the context of our example:- The algebraic fraction \(\frac{13}{3+\sqrt{11}}\) needed transformation.- Through multiplication by the conjugate and simplification, we obtained a manageable form: \[ -\frac{39}{2} + \frac{13\sqrt{11}}{2} \]

This form, where the denominator is rational, is often preferred because it's easier to interpret and work with in both algebraic equations and real-world applications.