Problem 381 - Elementary Algebra

Question

In the following exercises, simplify by rationalizing the denominator. $$ \frac{\sqrt{2}}{\sqrt{x}-\sqrt{3}} $$

Step-by-Step Solution

Answer

\( \frac{ \sqrt{2x} + \sqrt{6} }{ x - 3 } \)

Step 1 - Identify the Conjugate

To rationalize the denominator, identify the conjugate of the denominator, which is \( \sqrt{x} - \sqrt{3} \). The conjugate will be \( \sqrt{x} + \sqrt{3} \).

Step 2 - Multiply Numerator and Denominator by the Conjugate

Multiply both the numerator and the denominator by the conjugate \( \sqrt{x} + \sqrt{3} \) to get: \[ \frac{ \sqrt{2} ( \sqrt{x} + \sqrt{3} ) }{(\ \sqrt{x} - \sqrt{3} ) (\ \sqrt{x} + \sqrt{3} )}. \]

Step 3 - Expand the Numerator

Expand the numerator using the distributive property: \[ \sqrt{2} (\ \sqrt{x} + \sqrt{3} ) = \sqrt{2} \sqrt{x} + \sqrt{2} \sqrt{3} = \sqrt{2x} + \sqrt{6}. \]

Step 4 - Simplify the Denominator

Use the difference of squares formula to simplify the denominator: \[ (\ \sqrt{x} )^2 - (\ \sqrt{3} )^2 = x - 3. \]

Step 5 - Combine the Results

Combine the simplified numerator and denominator: \[ \frac{ \sqrt{2x} + \sqrt{6} }{ x - 3 }. \]

Key Concepts

Conjugate

Rationalizing the denominator is easier when we understand the concept of a conjugate. In simple terms, the conjugate of a binomial is formed by changing the sign between two terms. For example, the conjugate of \( \sqrt{x} - \sqrt{3} \) is \( \sqrt{x} + \sqrt{3} \).

We use the conjugate to eliminate radicals from the denominator. This method is effective because when you multiply any binomial by its conjugate, you get a difference of squares, which we explore more in the 'Difference of Squares' section.

Here's how to find the conjugate:

• Identify the two terms in the denominator.
• Change the sign between those two terms.

For instance, the given problem has \( \sqrt{x} - \sqrt{3} \) in the denominator. Its conjugate is \( \sqrt{x} + \sqrt{3} \). By multiplying the numerator and denominator by this conjugate, we can eliminate the square roots from the denominator.

Distributive Property

The distributive property is a fundamental algebraic principle. It states that a term outside the parenthesis can be multiplied to each term inside the parenthesis. In our example, the term outside the parenthesis is \( \sqrt{2} \), and the terms inside are \( \sqrt{x} + \sqrt{3} \).

When we multiply \( \sqrt{2} ( \sqrt{x} + \sqrt{3} ) \), the distributive property helps us to break it down into:
\[ \sqrt{2} \sqrt{x} + \sqrt{2} \sqrt{3} = \sqrt{2x} + \sqrt{6} \. \]
Simplifying the numerator using this property makes complex-looking expressions easier to manage.

Steps to use the distributive property effectively:

• Multiply the term outside the parenthesis by each term inside the parenthesis.
• Combine like terms if possible (though, in our case, there are no like terms).

Difference of Squares

The difference of squares is another important concept in algebra. When multiplying a binomial by its conjugate, you get what is called a difference of squares.

The formula to remember is: \[(a - b)(a + b) = a^2 - b^2 \].

In the denominator of our problem, we have \( ( \sqrt{x} - \sqrt{3} ) ( \sqrt{x} + \sqrt{3} )\), which fits this pattern:
\[ ( \sqrt{x} )^2 - ( \sqrt{3} )^2 \].

• \( a = \sqrt{x} \)
• \( b = \sqrt{3} \)

The result is:
\[ ( \sqrt{x} )^2 - ( \sqrt{3} )^2 = x - 3. \]
The conjugate eliminates the square roots, leaving a polynomial that is much simpler to work with.

By using the difference of squares method, complex radicals can be simplified effectively, making the rest of the problem easier to solve.