College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.4 - Quadratic Equations - 1.4 Exercises - Page 112: 22


$x=\left\{ \dfrac{2}{3} \right\}$

Work Step by Step

The two numbers whose product is $ac= 9(4)=36 $ and whose sum is $b= -12 $ are $\{ -6,-6 \}$. Using these two numbers to decompose the middle term of the given equation, $ 9x^2-12x+4=0 ,$ results to \begin{array}{l}\require{cancel} 9x^2-6x-6x+4=0 \\\\ (9x^2-6x)-(6x-4)=0 \\\\ 3x(3x-2)-2(3x-2)=0 \\\\ (3x-2)(3x-2)=0 .\end{array} Equating each factor to zero (or the Zero-Factor Property), the solutions to the given equation are \begin{array}{l}\require{cancel} 3x-2=0 \\\\ 3x=2 \\\\ x=\dfrac{2}{3} ,\\\\\text{OR}\\\\ 3x-2=0 \\\\ 3x=2 \\\\ x=\dfrac{2}{3} .\end{array} Hence, the solution is $ x=\left\{ \dfrac{2}{3} \right\} .$
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