Problem 29 - College Algebra with Corequisite Support

Question

For the following exercises, simplify each expression. \(\frac{8}{1-\sqrt{17}}\)

Step-by-Step Solution

Answer

The simplified expression is \(-\frac{1}{2} - \frac{\sqrt{17}}{2}\).

Step 1: Identify the expression to simplify

The expression we need to simplify is \( \frac{8}{1 - \sqrt{17}} \). Our goal is to eliminate the square root from the denominator.

Step 2: Recognize the need for rationalizing the denominator

To rationalize the denominator, we multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of \(1 - \sqrt{17}\) is \(1 + \sqrt{17}\).

Step 3: Multiply by the conjugate

Multiply both the numerator and denominator of the fraction by \(1 + \sqrt{17}\).This gives:\[\frac{8}{1 - \sqrt{17}} \times \frac{1 + \sqrt{17}}{1 + \sqrt{17}} = \frac{8(1 + \sqrt{17})}{(1 - \sqrt{17})(1 + \sqrt{17})}\]

Step 4: Simplify the denominator using the difference of squares

The denominator can be simplified using the difference of squares formula: \((a - b)(a + b) = a^2 - b^2\).This gives:\[(1 - \sqrt{17})(1 + \sqrt{17}) = 1^2 - (\sqrt{17})^2 = 1 - 17 = -16\]Thus, the expression now becomes:\[\frac{8(1 + \sqrt{17})}{-16}\]

Step 5: Simplify the fraction

Distribute the 8 in the numerator:\[8(1 + \sqrt{17}) = 8 + 8\sqrt{17}\]Then, divide each term in the numerator by \(-16\):\[\frac{8 + 8\sqrt{17}}{-16} = \frac{8}{-16} + \frac{8\sqrt{17}}{-16} = -\frac{1}{2} - \frac{\sqrt{17}}{2}\]

Step 6: Write the final simplified expression

Thus, the simplified expression is:\[-\frac{1}{2} - \frac{\sqrt{17}}{2}\]

Key Concepts

Difference of Squares

The difference of squares is a fundamental algebraic identity that can help simplify expressions and calculations. It's expressed as \[(a - b)(a + b) = a^2 - b^2\] This formula shows how the product of the difference and sum of two terms simplifies into the subtraction of their squares. In the case of simplifying the expression \[\frac{8}{1 - \sqrt{17}}\], we apply this identity when multiplying by the conjugate \(1 + \sqrt{17}\). This allows us to clear the square root from the denominator.

• In our case, \(a = 1\) and \(b = \sqrt{17}\).
• This leads to \((1 - \sqrt{17})(1 + \sqrt{17}) = 1^2 - (\sqrt{17})^2\).
• Simplified further: \(1 - 17 = -16\).

Using the difference of squares helps transform complex expressions into simpler forms.

Simplifying Expressions

Simplifying expressions is a crucial skill in algebra. It involves reducing expressions to their simplest form while retaining their value. In our exercise, we started with the fraction \(\frac{8}{1-\sqrt{17}}\). This expression initially seemed complex due to the square root in the denominator.To simplify, our goal was to remove the square root by rationalizing the denominator:

• This process involved multiplying by the conjugate, effectively using the difference of squares formula to simplify the denominator to \(-16\).
• After simplifying the denominator, the numerator \(8(1+\sqrt{17})\) was expanded to \(8 + 8\sqrt{17}\).
• Each part was then divided by \(-16\), reducing the whole expression to \[\frac{-1}{2} - \frac{\sqrt{17}}{2}\].

This created a cleaner, more straightforward expression that's easier to work with in further calculations.

Conjugate in Algebra

The conjugate in algebra is a tool used to eliminate square roots or complex numbers from denominators. It's particularly useful in rationalizing the denominator of expressions like \(\frac{8}{1 - \sqrt{17}}\). For any binomial expression \(a - b\), the conjugate is \(a + b\).In rationalizing denominators, the conjugate pairs are multiplied to exploit the difference of squares:

• Here, the original expression's denominator \(1 - \sqrt{17}\) was multiplied with its conjugate \(1 + \sqrt{17}\) to form a simple number, \(-16\).
• This action effectively removed irrational parts from the denominator.

It's not just about a neat trick. Using the conjugate simplifies and removes complexity, making further algebraic manipulation more straightforward. It's a powerful technique seen often in algebra, and crucial for working with irrational expressions.