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: 65



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ (x-9)(x-1)=-16 ,$ use the FOIL Method and express the equation in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} x(x)+x(-1)-9(x)-9(-1)=-16 \\\\ x^2-x-9x+9=-16 \\\\ x^2+(-x-9x)+(9+16)=0 \\\\ x^2-10x+25=0 .\end{array} In the equation above, $a= 1 ,$ $b= -10,$ and $c= 25 .$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then \begin{array}{l}\require{cancel} x=\dfrac{-(-10)\pm\sqrt{(-10)^2-4(1)(25)}}{2(1)} \\\\ x=\dfrac{10\pm\sqrt{100-100}}{2} \\\\ x=\dfrac{10\pm\sqrt{0}}{2} \\\\ x=\dfrac{10}{2} \\\\ x=5 .\end{array} The solution is $ x=5 .$
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