Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.5 - Solving Equations by the Zero-Factor Property - 5.5 Exercises: 25

Answer

$x=-\dfrac{4}{3}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ 9x^2+24x+16=0 ,$ in factored form. Then, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation. $\bf{\text{Solution Details:}}$ The factored form of the equation above is \begin{array}{l}\require{cancel} (3x+4)(3x+4)=0 .\end{array} Equating each factor to zero (Zero-Factor Property), then \begin{array}{l}\require{cancel} 3x+4=0 \text{ OR } 3x+4=0 .\end{array} Using the properties of equality to solve each of the equation above results to \begin{array}{l}\require{cancel} 3x+4=0 \\\\ 3x=-4 \\\\ x=-\dfrac{4}{3} \\\\\text{ OR }\\\\ 3x+4=0 \\\\ 3x=-4 \\\\ x=-\dfrac{4}{3} .\end{array} Hence, the solutions are $ x=-\dfrac{4}{3} .$ $\bf{\text{Supplementary Solution/s:}}$ To factor the quadratic expression $ax^2+bx+c,$ find two numbers whose product is $ac$ and whose sum is $b$. Use these $2$ numbers to decompose the middle term of the quadratic expression and then use factoring by grouping. In the expression, $ 9x^2+24x+16 ,$ the value of $ac$ is $ 9(16)=144 $ and the value of $b$ is $ 24 .$ The $2$ numbers that have a product $ac$ and a sum of $b$ are $\{ 12,12 \}.$ Using these $2$ numbers to decompose the middle term of the given expression results to \begin{array}{l}\require{cancel} 9x^2+12x+12x+16 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (9x^2+12x)+(12x+16) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3x(3x+4)+4(3x+4) .\end{array} Factoring the $GCF= (3x+4) $ of the entire expression above results to \begin{array}{l}\require{cancel} (3x+4)(3x+4) .\end{array}
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