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Ch. R - Review of Basic Concepts
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
All textbooksLial 13th EditionCh. R - Review of Basic ConceptsProblem 156
Chapter 1, Problem 156

Rationalize each denominator. Assume all variables represent nonnegative numbers and that no denominators are 0.
a(√a+b)−1\(\frac{a}{(\surd a+b)-1}\)

Verified step by step guidance
1
Identify the expression to rationalize: \(\frac{a}{\sqrt{a} + b} - 1\).
Focus on rationalizing the denominator of the fraction \(\frac{a}{\sqrt{a} + b}\). To do this, multiply both the numerator and denominator by the conjugate of the denominator, which is \(\sqrt{a} - b\).
Multiply numerator and denominator by \(\sqrt{a} - b\): \(\frac{a}{\sqrt{a} + b} \times \frac{\sqrt{a} - b}{\sqrt{a} - b} = \frac{a(\sqrt{a} - b)}{(\sqrt{a} + b)(\sqrt{a} - b)}\).
Simplify the denominator using the difference of squares formula: \((\sqrt{a} + b)(\sqrt{a} - b) = (\sqrt{a})^2 - b^2 = a - b^2\).
Rewrite the expression as \(\frac{a(\sqrt{a} - b)}{a - b^2} - 1\) and then combine the terms over a common denominator if needed.

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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 square roots or irrational numbers from the denominator of a fraction. This is done by multiplying the numerator and denominator by a conjugate or an appropriate expression to create a rational denominator, simplifying the expression.
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Guided course
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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 uses the difference of squares formula, which removes the square root terms in the denominator, making it rational and easier to work with.
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Complex Conjugates

Properties of Square Roots and Nonnegative Variables

Since variables represent nonnegative numbers, square roots like √a are defined and real. This assumption ensures that expressions involving square roots are valid and simplifies the process of rationalization without considering complex numbers.
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Imaginary Roots with the Square Root Property
Related Practice
Textbook Question

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