Question

Rewrite each expression by rationalizing the denominator. $$ \frac{10}{\sqrt{6}-1} $$

Short answer

Answer: The expression with the rationalized denominator is \(2(\sqrt{6}+1)\).

Step-by-step solution

Identify the conjugate of the denominator#

The conjugate of the denominator, \(\sqrt{6}-1\), is \(\sqrt{6}+1\). We will multiply both the numerator and denominator by this conjugate to rationalize the denominator.

Multiply by the conjugate#

Multiply both the numerator and denominator by the conjugate, \(\sqrt{6}+1\): $$ \frac{10}{\sqrt{6}-1} \cdot \frac{\sqrt{6}+1}{\sqrt{6}+1} $$

Distribute in the numerator and use the difference of squares in the denominator#

Now distribute \(10(\sqrt{6}+1)\) in the numerator and use the difference of squares formula in the denominator, which is \((a+b)(a-b) = a^2 - b^2\): $$ \frac{10\sqrt{6}+10}{(\sqrt{6})^2 - (1)^2} $$

Simplify the expression#

Simplify the expression by evaluating the squares and subtracting them in the denominator: $$ \frac{10\sqrt{6}+10}{6-1} $$ $$ \frac{10\sqrt{6}+10}{5} $$

Factor out the common factor in the numerator#

Factor out the common factor "10" in the numerator: $$ \frac{10(\sqrt{6}+1)}{5} $$

Reduce the common factors#

Reduce the common factors between the numerator and the denominator: $$ \frac{2(\sqrt{6}+1)}{1} $$ The expression with the rationalized denominator is: $$ 2(\sqrt{6}+1) $$

Key concepts

Conjugate

When you encounter expressions with square roots in the denominator, rationalizing the denominator is essential. This process often begins with the concept of a **conjugate**. The conjugate of a binomial, such as \(\sqrt{6} - 1\), is formed by changing the sign between the two terms, giving us \(\sqrt{6} + 1\).

Why do we use the conjugate? When we multiply the original expression by its conjugate, it helps eliminate the square root in the denominator. This is because the product of a binomial and its conjugate is known as a **difference of squares**, which results in a rational number.

• The conjugate changes the expression from complex to simple.
• It's a helpful tool for simplifying expressions involving radicals.
• Using the conjugate allows us to express the number in a form that’s easier to handle and interpret.

Difference of Squares

The concept of **difference of squares** is a significant algebraic tool used when rationalizing denominators. When we multiply an expression by its conjugate, the result is expressed as \((a + b)(a - b) = a^2 - b^2\). This formula is exactly what happens in the denominator.

For example, consider \((\sqrt{6} + 1)(\sqrt{6} - 1)\):

• Here, \(a = \sqrt{6}\) and \(b = 1\).
• The expression turns into \(a^2 - b^2 = (\sqrt{6})^2 - 1^2 = 6 - 1 = 5\).

In this way, the square root is removed, leaving us with a simple integer in the denominator.

The difference of squares not only simplifies the denominator but also makes the entire expression more straightforward, allowing easier manipulation and comparison.

Simplifying Expressions

**Simplifying expressions** involves reducing them to their simplest form while retaining the same value. After rationalizing the denominator using the conjugate and difference of squares, the next step in this process is simplification.

Consider our example:

• In the numerator, \(10(\sqrt{6} + 1)\) is expanded to \(10\sqrt{6} + 10\).
• Meanwhile, the denominator simplifies to 5.

To simplify further, factor out any common factors from the numerator and denominator. We notice that 10 is a common factor:

• Factoring gives us \(\frac{10(\sqrt{6} + 1)}{5}\).

Finally, reduce the expression by dividing the numerator and denominator by their greatest common divisor, which, in this case, is 5:

• This results in \(2(\sqrt{6} + 1)\).

This simplified expression keeps the mathematical integrity of the original while rendering it easier to use and understand.