Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 4 - Quadratic Functions - 4.5 Solving Equations by Factoring - 4.5 Exercises - Page 359: 26


$\color{blue}{\left\{-2.5, 2.5\right\}}$

Work Step by Step

Subtract $5$ to both sides: $4x^2-20-5=5-5 \\4x^2-25=0 \\(2x)^2-5^2=0$ Factor the binomial using the formula $a^2-b^2=(a-b)(a+b)$, where $a=2x$ and $b=5$ to obtain: $(2x-5)(2x+5)=0$ RECALL: The Zero-Factor Property states that if $ab=0$, then $a=0$ or $b=0$, or both are zero. Use the Zero-Factor Property by equating each factor to zero to obtain: $2x-5=0$ or $2x+5=0$ Solve each equation to obtain: $2x-5=0 \\2x=5 \\\frac{2x}{2}=\frac{5}{2} \\x=2.5$ or $2x+5=0 \\2x+5-5=0-5 \\2x=-5 \\\frac{2x}{2}=\frac{-5}{2} \\x=-2.5$ Therefore, the solution set is $\color{blue}{\left\{-2.5, 2.5\right\}}$.
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