Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises: 39


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

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given eqution, $ x^2-5x+6=0 ,$ express the equation in factored form. Then equate each factor to zero (Zero Product Property). Finally, solve each equation. $\bf{\text{Solution Details:}}$ In the trinomial expression above, the value of $c$ is $ 6 $ and the value of $b$ is $ -5 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,6 \}, \{ 2,3 \}, \\ \{ -1,-6 \}, \{ -2,-3 \} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -2,-3 \}.$ Hence, the factored form of the equation above is \begin{array}{l}\require{cancel} (x-2)(x-3)=0 .\end{array} Equating each factor to zero (Zero Product Property), the solutions to the equation above are \begin{array}{l}\require{cancel} x-2=0 \\\\\text{OR}\\\\ x-3=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-2=0 \\\\ x=2 \\\\\text{OR}\\\\ x-3=0 \\\\ x=3 .\end{array} Hence, $ x=\left\{ 2,3 \right\} .$
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