Chapter 6: Problem 33
In Exercises \(33-37\), rewrite each expression by rationalizing the denominator. $$ \frac{2}{\sqrt{3}+1} $$
Short Answer
Expert verified
Question: Rewrite the given expression with rationalized denominator: $$\frac{2}{\sqrt{3} + 1}$$
Answer: $$\sqrt{3} - 1$$
Step by step solution
01
Identify the conjugate of the denominator.
The conjugate of the denominator \((\sqrt{3} + 1)\) is \((\sqrt{3} - 1)\). We will multiply the given fraction by \(\frac{\sqrt{3} - 1}{\sqrt{3} -1}\). This is a clever form of 1, as it does not change the value of the original fraction but will help us remove the square root from the denominator.
02
Multiply the given fraction by the conjugate.
We multiply both the numerator and denominator of the given fraction by the conjugate of the denominator. So, we have:
$$
\frac{2}{\sqrt{3} + 1} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} -1}
$$
03
Simplify the numerator and denominator.
We perform the multiplication and simplify the numerator and the denominator:
$$
\frac{2(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} -1)}
$$
Expanding and combining like terms, we get:
$$
\frac{2\sqrt{3} - 2}{(\sqrt{3})^2 - 1^2}
$$
04
Further simplify and write the final answer.
Simplify the expression by performing the operations in the numerator and denominator:
$$
\frac{2\sqrt{3} - 2}{3 - 1}
$$
$$
\frac{2\sqrt{3} - 2}{2}
$$
Now, we can factor out a 2 from the numerator and cancel the 2's in the numerator and the denominator:
$$
\frac{2(\sqrt{3} - 1)}{2} = \sqrt{3} - 1
$$
So the given expression with rationalized denominator is \(\sqrt{3} - 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Conjugate
When dealing with algebraic expressions that have square roots in the denominator, itβs common to use the concept of a conjugate. The conjugate of a binomial expression like \( a + b \) is \( a - b \). This involves changing the sign of the second term.
The reason we use conjugates in the context of rationalizing denominators is to eliminate radicals or square roots from the denominator.
The reason we use conjugates in the context of rationalizing denominators is to eliminate radicals or square roots from the denominator.
- For \( \sqrt{3} + 1 \), the conjugate is \( \sqrt{3} - 1 \).
- The product of a binomial and its conjugate results in a difference of squares: \((a + b)(a - b) = a^2 - b^2\).
Steps to Simplification
Simplification is the process where we take a complex expression and reduce it to its simplest form. This often involves factoring, combining like terms, or performing basic arithmetic operations.
The simplification process in rationalizing the denominator is crucial because it allows the fraction to be expressed in a more standard form. Here's how you do it:
The simplification process in rationalizing the denominator is crucial because it allows the fraction to be expressed in a more standard form. Here's how you do it:
- Multiply both the numerator and the denominator by the conjugate of the denominator.
- Perform the multiplication step by expanding both the numerator and the denominator.
- Apply the difference of squares formula to the denominator: convert \((\sqrt{3} + 1)(\sqrt{3} - 1)\) into \( (\sqrt{3})^2 - 1^2 \).
- Simplify both the numerator and the denominator until no further simplification is possible.
Exploring Algebraic Expressions
Algebraic expressions are combinations of variables, numbers, and at times operations like addition or multiplication. They form the basis of algebra, where finding equivalent forms can simplify problem-solving.
In the exercise provided, the algebraic expression \( \frac{2}{\sqrt{3}+1} \) involves a fraction where the denominator is a binomial expression with a square root.
In the exercise provided, the algebraic expression \( \frac{2}{\sqrt{3}+1} \) involves a fraction where the denominator is a binomial expression with a square root.
- These expressions are manipulated using algebraic techniques like finding conjugates for rationalization.
- They highlight the importance of operations, such as multiplying terms and recognizing patterns like the difference of squares.
- Understanding these expressions enables students to tackle a wide range of mathematical problems by applying consistent rules and formulas.