Answer
$x=\left\{ 3-\sqrt{17},3+\sqrt{17}
\right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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\}
.$
