Answer
$x=\left\{ \dfrac{2-i}{4},\dfrac{2+i}{4} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
(2x-1)(8x-4)=-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}
2x(8x)+2x(-4)-1(8x)-1(-4)=-1
\\\\
16x^2-8x-8x+4=-1
\\\\
16x^2-16x+4=-1
\\\\
16x^2-16x+4+1=0
\\\\
16x^2-16x+5=0
.\end{array}
The quadratic equation above has $a=
16
, b=
-16
, c=
5
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
x=\dfrac{-(-16)\pm\sqrt{(-16)^2-4(16)(5)}}{2(16)}
\\\\
x=\dfrac{16\pm\sqrt{256-320}}{32}
\\\\
x=\dfrac{16\pm\sqrt{-64}}{32}
.\end{array}
Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
x=\dfrac{16\pm\sqrt{-1}\cdot\sqrt{64}}{32}
.\end{array}
Since $i=\sqrt{-1},$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
x=\dfrac{16\pm i\sqrt{64}}{32}
.\end{array}
Simplifying the radicand by writing it as an expression that contains a factor that is a perfect square of the index and then extracting the root of that factor, the equation above is equivalent to
\begin{array}{l}\require{cancel}
x=\dfrac{16\pm i\sqrt{(8)^2}}{32}
\\\\
x=\dfrac{16\pm 8i}{32}
.\end{array}
Cancelling the common factor in each term results to
\begin{array}{l}\require{cancel}
x=\dfrac{\cancel8(2)\pm \cancel8(1)i}{\cancel8(4)}
\\\\
x=\dfrac{2\pm i}{4}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=\dfrac{2-i}{4}
\\\\\text{OR}\\\\
x=\dfrac{2+i}{4}
.\end{array}
Hence, $
x=\left\{ \dfrac{2-i}{4},\dfrac{2+i}{4} \right\}
.$
