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Ch. P - Fundamental Concepts of Algebra
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 1, Problem 53

Rationalize the denominator.
65+3\(\frac{6}{\sqrt5+\sqrt3}\)

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1
Identify the expression to rationalize: \(\frac{6}{\sqrt{5} + \sqrt{3}}\). The goal is to eliminate the square roots from the denominator.
Multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of \(\sqrt{5} + \sqrt{3}\) is \(\sqrt{5} - \sqrt{3}\). So multiply by \(\frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} - \sqrt{3}}\).
Apply the multiplication: The numerator becomes \(6(\sqrt{5} - \sqrt{3})\), and the denominator becomes \((\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3})\).
Use the difference of squares formula for the denominator: \((a + b)(a - b) = a^2 - b^2\). Here, \(a = \sqrt{5}\) and \(b = \sqrt{3}\), so the denominator simplifies to \$5 - 3$.
Write the new expression with the simplified denominator and the expanded numerator: \(\frac{6(\sqrt{5} - \sqrt{3})}{5 - 3}\). This expression has a rationalized denominator.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Rationalizing the Denominator

Rationalizing the denominator involves eliminating any radicals (square roots) from the denominator of a fraction. This is done to simplify the expression and make it easier to work with. Typically, this is achieved by multiplying the numerator and denominator by a conjugate or an appropriate radical expression.
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Conjugates of Binomials

The conjugate of a binomial expression like (√a + √b) is (√a - √b). Multiplying a binomial by its conjugate results in a difference of squares, which eliminates the square roots. This property is essential for rationalizing denominators containing sums or differences of square roots.
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Difference of Squares Formula

The difference of squares formula states that (x + y)(x - y) = x² - y². When applied to radicals, it helps remove square roots by converting the product into a subtraction of the radicands. This formula is key to simplifying expressions after multiplying by the conjugate.
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