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Ch. 1 - Equations and Inequalities
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
Chapter 2, Problem 81

Solve each equation. (2x+3)2/3 + (2x+3)1/3 - 6 = 0

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1
Identify the substitution to simplify the equation. Let \( y = (2x + 3)^{1/3} \). This means \( y^2 = (2x + 3)^{2/3} \).
Rewrite the original equation \( (2x+3)^{2/3} + (2x+3)^{1/3} - 6 = 0 \) in terms of \( y \) as \( y^2 + y - 6 = 0 \).
Solve the quadratic equation \( y^2 + y - 6 = 0 \) using factoring, completing the square, or the quadratic formula.
After finding the values of \( y \), substitute back \( y = (2x + 3)^{1/3} \) to get equations of the form \( (2x + 3)^{1/3} = y \).
Solve each resulting equation for \( x \) by cubing both sides to eliminate the cube root, then isolate \( x \).

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Key Concepts

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

Exponents and Rational Powers

Rational exponents represent roots and powers simultaneously, where a fractional exponent like a^(m/n) means the nth root of a raised to the mth power. Understanding how to manipulate expressions with fractional exponents is essential for simplifying and solving equations involving terms like (2x+3)^(2/3).
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Substitution Method for Solving Equations

Substitution involves replacing a complex expression with a single variable to simplify the equation. In this problem, letting y = (2x+3)^(1/3) transforms the equation into a quadratic form, making it easier to solve for y before back-substituting to find x.
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Solving Quadratic Equations

Quadratic equations are polynomial equations of degree two and can be solved using factoring, completing the square, or the quadratic formula. After substitution, the equation becomes quadratic in form, so knowing how to solve quadratics is crucial to finding the values of the substituted variable.
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