Chapter 10: Problem 35
Rationalize each denominator. See Example 4. $$ \frac{6}{2-\sqrt{7}} $$
Short Answer
Expert verified
The rationalized expression is \(-4 - 2\sqrt{7}\).
Step by step solution
01
Identify the Conjugate
To rationalize the denominator, first identify the conjugate of the denominator. The denominator is \(2 - \sqrt{7}\). The conjugate is \(2 + \sqrt{7}\).
02
Multiply by the Conjugate
Multiply both the numerator and the denominator by the conjugate. This gives:\[\frac{6}{2 - \sqrt{7}} \times \frac{2 + \sqrt{7}}{2 + \sqrt{7}} = \frac{6(2 + \sqrt{7})}{(2 - \sqrt{7})(2 + \sqrt{7})}\]
03
Simplify the Numerator
Distribute the \(6\) into the terms \(2\) and \(\sqrt{7}\):\[6 \times 2 + 6 \times \sqrt{7} = 12 + 6\sqrt{7}\]
04
Simplify the Denominator Using the Difference of Squares
The product \((2 - \sqrt{7})(2 + \sqrt{7})\) is a difference of squares, which is equal to:\[(2)^2 - (\sqrt{7})^2 = 4 - 7 = -3\]
05
Write the Simplified Expression
Combine the simplified numerator and denominator:\[\frac{12 + 6\sqrt{7}}{-3}\]
06
Simplify the Entire Fraction
Divide both terms in the numerator by \(-3\):\[\frac{12}{-3} + \frac{6\sqrt{7}}{-3} = -4 - 2\sqrt{7}\]
07
Final Answer
Thus, the rationalized form of the expression is:\(-4 - 2\sqrt{7}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Conjugate
The concept of a conjugate is crucial when it comes to rationalizing denominators, especially those containing square roots. A conjugate involves changing the sign between two terms in a binomial. For instance, consider the expression \(2 - \sqrt{7}\). Its conjugate would be \(2 + \sqrt{7}\).
Using the conjugate helps to eliminate square roots in the denominator of a fraction by leveraging the difference of squares. By multiplying the numerator and the denominator of a fraction by the conjugate of the denominator, it transforms the denominator into a simpler expression without square roots.
This technique simplifies calculation and makes the expression more accessible by ensuring the denominator is a rational number.
Using the conjugate helps to eliminate square roots in the denominator of a fraction by leveraging the difference of squares. By multiplying the numerator and the denominator of a fraction by the conjugate of the denominator, it transforms the denominator into a simpler expression without square roots.
This technique simplifies calculation and makes the expression more accessible by ensuring the denominator is a rational number.
Difference of Squares
The difference of squares is a powerful algebraic identity that states \(a^2 - b^2 = (a - b)(a + b)\). It is particularly useful when rationalizing denominators that include square roots.
When you identify a denominator such as \(2 - \sqrt{7}\) and multiply it by its conjugate, \(2 + \sqrt{7}\), you get a difference of squares:
When you identify a denominator such as \(2 - \sqrt{7}\) and multiply it by its conjugate, \(2 + \sqrt{7}\), you get a difference of squares:
- First, compute \((2)^2\), which is 4.
- Next, compute \((\sqrt{7})^2\), which is 7.
- Subtract these results to get \(4 - 7 = -3\).
Square Roots
Square roots often pose a challenge when found in denominators of fractions. Simplifying expressions with square roots involves using conjugates, which transform problems with roots into rational numbers.
A square root, like \(\sqrt{7}\), represents a number which, when squared, returns the original number 7. In processes like rationalizing denominators, these square roots can appear alongside other numbers in binomials, like \(2 - \sqrt{7}\).
By multiplying this by its conjugate, the square roots cancel out (thanks to the difference of squares), resulting in a number without any roots in the denominator. This simplifies expressions greatly and allows for clearer and more concise answers.
A square root, like \(\sqrt{7}\), represents a number which, when squared, returns the original number 7. In processes like rationalizing denominators, these square roots can appear alongside other numbers in binomials, like \(2 - \sqrt{7}\).
By multiplying this by its conjugate, the square roots cancel out (thanks to the difference of squares), resulting in a number without any roots in the denominator. This simplifies expressions greatly and allows for clearer and more concise answers.
Fractions
Understanding fractions and how to manipulate them is a cornerstone of algebra that applies to many topics, including rationalizing denominators. A fraction consists of a numerator and a denominator, and in cases where the denominator contains square roots, it is necessary to rationalize it for simplicity.
Consider the fraction \(\frac{6}{2 - \sqrt{7}}\). By multiplying both the numerator and the denominator by the conjugate of the denominator, \(2 + \sqrt{7}\), we effectively "remove" the square root from the denominator.
This results in a simpler expression. The fraction \(\frac{6(2 + \sqrt{7})}{-3}\) can then be further simplified by dividing each term of the numerator by \(-3\), leading to the final answer \(-4 - 2\sqrt{7}\). This process is essential for expressing fractions in their simplest form.
Consider the fraction \(\frac{6}{2 - \sqrt{7}}\). By multiplying both the numerator and the denominator by the conjugate of the denominator, \(2 + \sqrt{7}\), we effectively "remove" the square root from the denominator.
This results in a simpler expression. The fraction \(\frac{6(2 + \sqrt{7})}{-3}\) can then be further simplified by dividing each term of the numerator by \(-3\), leading to the final answer \(-4 - 2\sqrt{7}\). This process is essential for expressing fractions in their simplest form.