Problem 2 - Algebra 2 and Trigonometry

Question

To rationalize the denominator of \(\frac{4}{2+\sqrt{8}},\) Brittany multiplied by \(\frac{2-\sqrt{8}}{2}\) and Justin multiplied by \(\frac{1-\sqrt{2}}{1-\sqrt{2}} .\) Explain why both are correct.

Step-by-Step Solution

Answer

Both methods aim to rationalize the denominator. Justin's is correct and conventional; Brittany's method is conceptually right but uses an incorrect multiplier.

Step 1: Understand the Problem

We need to rationalize the denominator of the fraction \( \frac{4}{2+\sqrt{8}} \). Rationalizing the denominator generally means removing any square roots from the denominator.

Step 2: Brittany's Method Explanation

Brittany multiplies the fraction by \( \frac{2-\sqrt{8}}{2} \). We aim to rationalize the denominator by using a different conjugate form, but Brittany's method is slightly incorrect as she should be using \( \frac{2-\sqrt{8}}{2-\sqrt{8}} \) as the multiplier to correctly create a difference of squares in the denominator.

Step 3: Justin's Method Explanation

Justin multiplies by the conjugate \( \frac{1-\sqrt{2}}{1-\sqrt{2}} \), which simplifies the denominator due to the identity \((a+b)(a-b) = a^2 - b^2\). This method is correct and will remove the square root from the denominator when simplified properly.

Step 4: Simplification Results

For Brittany's method, multiplying by \( \frac{2-\sqrt{8}}{2-\sqrt{8}} \) would create the result: \[ \frac{4(2-\sqrt{8})}{(2)^2-(\sqrt{8})^2} = \frac{8-4\sqrt{8}}{4-8} = \frac{8-4\sqrt{8}}{-4} \]. For Justin's method, multiplying by \( \frac{1-\sqrt{2}}{1-\sqrt{2}} \) yields \[ \frac{4(1-\sqrt{2})}{1-2} = \frac{4-4\sqrt{2}}{-1} \], both resulting in a rational denominator.

Step 5: Conclusion

Both methods achieve rationalization of the denominator, albeit Justin's method is more straightforward and conventional, using the proper conjugate form. Brittany's incorrect multiplier did not follow the principle of multiplying by the conjugate exactly, but she was on a correct path possibly through a mistaken calculation.

Key Concepts

Conjugate Method

The conjugate method is a powerful technique for rationalizing the denominator of a fraction, especially when dealing with expressions that contain square roots. The idea is to eliminate the square root by multiplying the numerator and denominator by the conjugate of the denominator.

To understand how the conjugate works, consider an expression like \( a + \sqrt{b} \). Its conjugate would be \( a - \sqrt{b} \). When these two expressions are multiplied together, the denominator becomes a difference of squares, which naturally simplifies to an expression without any square roots: \[(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b\]

This difference of squares formula helps in eliminating the square root from the denominator. In essence, Brittany initially aimed to multiply by the conjugate \( 2 - \sqrt{8} \), which was a step in the right direction. However, she missed using the full conjugate to keep the equation balanced. Justin succeeded with \( 1 - \sqrt{2} \), since this accurate application of the conjugate method ensured the radicals were removed effectively.

Difference of Squares

The difference of squares is a critical algebraic identity that is frequently used when multiplying conjugates to rationalize the denominator. The formula is expressed as \[a^2 - b^2 = (a+b)(a-b)\]

This relation showcases how multiplying a sum and a difference of the same terms results in a simple subtraction of squares, allowing any square roots to cancel out.

Let's see this in action by applying Brittany's case (using the correct conjugate). When attempting to rationalize \( 2 + \sqrt{8} \), the product with \( 2 - \sqrt{8} \) creates:

• \( 2^2 - (\sqrt{8})^2 \)
• Simplifying gives the difference: \( 4 - 8 = -4 \)

Justin applied the concept using \( 1 + \sqrt{2} \), ending with a negative but whole number denominator which proves the usefulness of identifying difference of squares during rationalization.

Simplifying Radicals

Simplifying radicals involves expressing numbers under the square root in their simplest form. It's crucial for breaking down complex square roots before rationalizing denominators.

In processing radicals, consider \( \sqrt{8} \), which breaks down to \( \sqrt{4 \times 2} = 2\sqrt{2} \). This factorization simplifies subsequent operations. Both Brittany and Justin worked with rewritten radicals during their calculations.

Before taking any further steps to rationalize, simplifying radicals ensures simpler and cleaner algebraic manipulations. Follow these steps to simplify a radical:

• Identify perfect square factors within the radical.
• Express the radical as the product of the squares and the leftover factor.
• Simplify by taking the square root of the perfect squares.

The simplification process aids in clear, manageable expressions, easing the way for subsequent operations involving conjugates and difference of squares.