Problem 372 - Elementary Algebra

Question

In the following exercises, simplify by rationalizing the denominator. $$ \frac{\sqrt{s}-\sqrt{6}}{\sqrt{s}+\sqrt{6}} $$

Step-by-Step Solution

Answer

The simplified form is \frac{s - 2 \sqrt{6s} + 6}{s - 6}\

Step 1: Identify the Conjugate

The conjugate of the denominator \( \sqrt{s} + \sqrt{6} \ \) is \( \sqrt{s} - \sqrt{6} \ \). We will use this to rationalize the denominator.

Step 2: Multiply Numerator and Denominator by the Conjugate

Multiply the numerator and the denominator by \sqrt{s} - \sqrt{6}\ to get: \[ \frac{( \sqrt{s} - \sqrt{6})( \sqrt{s} - \sqrt{6})}{( \sqrt{s} + \sqrt{6})( \sqrt{s} - \sqrt{6})}\ \]

Step 3: Apply the Difference of Squares

Use the difference of squares formula \( (a+b)(a-b) = a^2 - b^2 \) in the denominator: \[ ( \sqrt{s} + \sqrt{6})( \sqrt{s} - \sqrt{6}) = s - 6 \]

Step 4: Expand the Numerator

Expand the numerator \( ( \sqrt{s} - \sqrt{6})( \sqrt{s} - \sqrt{6}) \): \[ \sqrt{s} \times \sqrt{s} - 2 \sqrt{s} \sqrt{6} + \sqrt{6} \sqrt{6} \ = s - 2 \sqrt{6s} + 6 \]

Step 5: Simplify the Expression

Combine the expanded terms to write the fraction: \[ \frac{s - 2 \sqrt{6s} + 6}{s - 6}\ \]

Key Concepts

Simplifying Expressions

Simplifying expressions means making them simpler and easier to understand or work with. It often requires reducing expressions to their most basic form.
For instance, in the given problem, we simplified the fraction by rationalizing the denominator.
This process ensures the denominator is no longer an irrational number.

Here’s the breakdown:

• First, identify the conjugate of the denominator to eliminate the square roots.
• Then multiply both the numerator and the denominator by this conjugate.
• After multiplying, use algebraic identities to simplify further.

A simplified expression is always more manageable and often reveals important features of the problem.

Conjugates

A conjugate in math is a pair of expressions that differ only in the sign between two terms.
For real numbers, it’s often used to help with rationalizing denominators. In this problem, the conjugate of the denominator \( \sqrt{s} + \sqrt{6} \) is \( \sqrt{s} - \sqrt{6} \).

Conjugates have a special property: when you multiply a number by its conjugate, you get a difference of squares. This is helpful to remove irrational numbers in the denominator.

For example:

• If \( a + b \) is a binomial, its conjugate is \( a - b \).
• Multiplying \( a + b \) and \( a - b \) results in \( a^2 - b^2 \). This eliminates the middle term, simplifying the expression.

In practice, this transforms a difficult expression into something more straightforward and easier to handle.

Difference of Squares

The difference of squares is a fundamental algebraic identity, expressed as:
\( a^2 - b^2 = (a + b)(a - b) \).

It means that the product of the sum and difference of two terms is equal to the difference of their squares.
In rationalizing denominators, this property helps eliminate square roots.

Here’s how it works in our problem:

• First, multiply the numerator and denominator by the conjugate \( \sqrt{s} - \sqrt{6} \).
• This yields a denominator which is \( (\sqrt{s} + \sqrt{6})(\sqrt{s} - \sqrt{6}) \).

Applying the difference of squares identity:

• The denominator simplifies to \( s - 6 \), since \( (\sqrt{s})^2 = s \) and \( (\sqrt{6})^2 = 6 \).

Using this step helps to remove square roots in denominators, making the expression more standard and easier to use in further calculations.