Problem 67 - College Algebra with Corequisite Support

Question

For the following exercises, simplify each expression. \(\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}}-2^{\frac{1}{2}}\)

Step-by-Step Solution

Answer

The simplified expression is \( \frac{-5\sqrt{2} - 6}{7} \).

Step 1: Simplify individual square roots

The expression \( \frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}}-2^{\frac{1}{2}} \) can be broken down by first simplifying \( \sqrt{8} \) and \( \sqrt{16} \). We have: 1. \( \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \), because \( \sqrt{4} = 2 \).2. \( \sqrt{16} = 4 \), because \( 16 \) is a perfect square.

Step 2: Substitute simplified square roots into expression

Substitute the simplified forms of \( \sqrt{8} \) and \( \sqrt{16} \) back into the expression: \( \frac{\sqrt{8} - \sqrt{16}}{4 - \sqrt{2}} - 2^{\frac{1}{2}} = \frac{2\sqrt{2} - 4}{4 - \sqrt{2}} - 2^{\frac{1}{2}} \).

Step 3: Simplify fraction numerator

To simplify the numerator \( 2\sqrt{2} - 4 \), note that it cannot be simplified further as it already consists of separate terms \( 2\sqrt{2} \) and \(-4 \).

Step 4: Rationalize the denominator

Rationalize the denominator \( 4 - \sqrt{2} \) by multiplying both the numerator and denominator by the conjugate \( 4 + \sqrt{2} \):\[\frac{(2\sqrt{2} - 4)(4 + \sqrt{2})}{(4 - \sqrt{2})(4 + \sqrt{2})}\]The denominator becomes \( 4^2 - (\sqrt{2})^2 = 16 - 2 = 14 \).

Step 5: Expand the numerator

Expand the numerator using distributive property:\( (2\sqrt{2} - 4)(4 + \sqrt{2}) = 2\sqrt{2} \cdot 4 + 2\sqrt{2} \cdot \sqrt{2} - 4 \cdot 4 - 4 \cdot \sqrt{2} \).This simplifies to:\( 8\sqrt{2} + 4 - 16 - 4\sqrt{2} = 4\sqrt{2} - 12 \).

Step 6: Simplify the main expression

Simplify the fraction \( \frac{4\sqrt{2} - 12}{14} \) by splitting it into two parts:\[\frac{4\sqrt{2}}{14} + \frac{-12}{14} = \frac{2\sqrt{2}}{7} - \frac{6}{7}\].So, the expression becomes:\( \frac{2\sqrt{2}}{7} - \frac{6}{7} - 2^{\frac{1}{2}} \).

Step 7: Simplify the expression further

Recognize that \( 2^{\frac{1}{2}} = \sqrt{2} \).Combine like terms:\( \frac{2\sqrt{2}}{7} - \sqrt{2} - \frac{6}{7} \). Convert \( \sqrt{2} \) to a common denominator:\( \sqrt{2} = \frac{7\sqrt{2}}{7} \).Combine with \( \frac{2\sqrt{2}}{7} - \sqrt{2} = \frac{2\sqrt{2} - 7\sqrt{2}}{7} = \frac{-5\sqrt{2}}{7} \).

Step 8: Final expression

Evaluate the final expression including all parts:\( \frac{-5\sqrt{2}}{7} - \frac{6}{7} = \frac{-5\sqrt{2} - 6}{7} \).This is the simplified form of the given expression.

Key Concepts

Square Roots

Square roots are mathematical functions that "un-do" squaring a number. In simpler terms, taking the square root of a number is finding a value that, when multiplied by itself, gives the original number. For instance, the square root of 16 is 4 because 4 multiplied by 4 equals 16. However, not all numbers are perfect squares like 16 and require simplification.

• To simplify, look for perfect square factors. For example, in the case of \( \sqrt{8} \), notice \( 8 = 4 \times 2 \).
• Since \( \sqrt{4} = 2 \), it gets simplified to \( 2\sqrt{2} \).

Dealing with square roots often involves breaking them down into simpler or smaller components, especially when they are part of a larger algebraic expression.

Rationalizing the Denominator

Rationalizing the denominator is a method used to eliminate square roots from the denominator of a fraction. This is crucial for simplifying expressions and making them easier to work with. To rationalize, multiply both the numerator and the denominator by the conjugate of the denominator:

• For example, for \( \frac{1}{4 - \sqrt{2}} \), the conjugate is \( 4 + \sqrt{2} \).
• Multiply both top and bottom: \( \frac{(4 + \sqrt{2})}{(4 - \sqrt{2})(4 + \sqrt{2})} \).

The denominator simplifies using the difference of squares: \[(4)^2 - (\sqrt{2})^2 = 16 - 2 = 14.\]
This process results in a "cleaner" or more simplified form of the fraction, without any irrational numbers in the denominator.

Common Denominator

A common denominator allows fractions to be added or subtracted by providing a common base for comparison. This is especially important when working with expressions involving multiple fractions or terms with roots.

• Consider \( \sqrt{2} \) as \( \frac{7\sqrt{2}}{7} \) to match the existing denominator.
• This conversion aids in combining terms like \( \frac{2\sqrt{2}}{7} \) and \( \frac{7\sqrt{2}}{7} \).

Such a conversion ensures the expressions can be added or subtracted directly, leading to effective simplification. The final expression maintains consistency, making calculations straightforward.

Distributive Property

The distributive property is a fundamental principle in algebra that helps in expanding or simplifying expressions. It states that a term multiplied by a sum is the same as the term multiplied by each addend individually. Therefore, for an expression like \((a + b)c\), it distributes to \(ac + bc\).

• For example, to simplify \((2\sqrt{2} - 4)(4 + \sqrt{2})\), apply the property: \(2\sqrt{2} \cdot 4 + 2\sqrt{2} \cdot \sqrt{2} - 4 \cdot 4 - 4 \cdot \sqrt{2}\).
• This results in breaking down the terms into \(8\sqrt{2} + 4 - 16 - 4\sqrt{2}\).

It helps combine like terms with similar radical expressions, making complex expressions easier to manage.