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: 13

Answer

$x=\{ 2, 3 \}$

Work Step by Step

The two numbers whose product is $ac=1(6)=6 $ and whose sum is $b= -5 $ are $\{ -2,-3 \}$. Using these two numbers to decompose the middle term of the given equation, $ x^2-5x+6=0 ,$ results to \begin{array}{l}\require{cancel} x^2-2x-3x+6=0 \\\\ (x^2-2x)-(3x-6)=0 \\\\ x(x-2)-3(x-2)=0 \\\\ (x-2)(x-3)=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} x-2=0 \\\\ x=2 ,\\\\\text{OR}\\\\ x-3=0 \\\\ x=3 .\end{array} Hence, the solutions are $ x=\{ 2, 3 \} .$
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