## Intermediate Algebra (12th Edition)

$x=\left\{ 3-\sqrt{17},3+\sqrt{17} \right\}$
$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $(x+1)(x-7)=1 ,$ express first 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(-7)+1(x)+1(-7)=1 \\\\ x^2-7x+x-7=1 \\\\ x^2-6x-7=1 .\end{array} Using the properties of equality, in the form $ax^2+bx+c=0,$ the expression above is equivalent to \begin{array}{l}\require{cancel} x^2-6x-7-1=0 \\\\ x^2-6x-8=0 .\end{array} The quadratic equation above has $a= 1 , b= -6 , c= -8 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-(-6)\pm\sqrt{(-6)^2-4(1)(-8)}}{2(1)} \\\\ x=\dfrac{6\pm\sqrt{36+32}}{2} \\\\ x=\dfrac{6\pm\sqrt{68}}{2} .\end{array} Simplifying the radical by writing the radicand as an expression that contains a factor that is a perfect power of the index and then extracting the root of that factor result to \begin{array}{l}\require{cancel} x=\dfrac{6\pm\sqrt{4\cdot17}}{2} \\\\ x=\dfrac{6\pm\sqrt{(2)^2\cdot17}}{2} \\\\ x=\dfrac{6\pm2\sqrt{17}}{2} .\end{array} Cancelling the common factors from all the terms results to \begin{array}{l}\require{cancel} x=\dfrac{\cancel2(3)\pm\cancel2(1)\sqrt{17}}{\cancel2(1)} \\\\ x=3\pm\sqrt{17} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=3-\sqrt{17} \\\\\text{OR}\\\\ x=3+\sqrt{17} .\end{array} Hence, $x=\left\{ 3-\sqrt{17},3+\sqrt{17} \right\} .$