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Rewrite each expression without absolute value bars. |12 - π|
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Ch. P - Fundamental Concepts of Algebra
Blitzer8th EditionCollege AlgebraISBN: 9780136970514
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TOC.tableOfContents
- Ch. P - Fundamental Concepts of Algebra311
- Ch. 1 - Equations and Inequalities398
- Ch. 2 - Functions and Graphs407
- Ch. 3 - Polynomial and Rational Functions306
- Ch. 4 - Exponential and Logarithmic Functions247
- Ch. 5 - Systems of Equations and Inequalities173
- Ch. 6 - Matrices and Determinants143
- Ch. 7 - Conic Sections124
- Ch. 8 - Sequences, Induction, and Probability251
Capitolo 1, Problema 53
Rationalize the denominator.
Guida verificata passo dopo passo1
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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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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Rationalizing Denominators
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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05:33Complex Conjugates
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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04:14Special Products - Square Formulas
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