Answer
$x=\left\{ \dfrac{-1-\sqrt{2}}{2},\dfrac{-1+\sqrt{2}}{2} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
4x^2+4x-1=0
,$ express in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula.
$\bf{\text{Solution Details:}}$
The quadratic equation above has $a=
4
, b=
4
, c=
-1
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
x=\dfrac{-4\pm\sqrt{4^2-4(4)(-1)}}{2(4)}
\\\\
x=\dfrac{-4\pm\sqrt{16+16}}{8}
\\\\
x=\dfrac{-4\pm\sqrt{32}}{8}
.\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{-4\pm\sqrt{16\cdot2}}{8}
\\\\
x=\dfrac{-4\pm\sqrt{(4)^2\cdot2}}{8}
\\\\
x=\dfrac{-4\pm4\sqrt{2}}{8}
.\end{array}
Cancelling the common factors from all the terms results to
\begin{array}{l}\require{cancel}
x=\dfrac{\cancel4(-1)\pm\cancel4(1)\sqrt{2}}{\cancel4(2)}
\\\\
x=\dfrac{-1\pm\sqrt{2}}{2}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=\dfrac{-1-\sqrt{2}}{2}
\\\\\text{OR}\\\\
x=\dfrac{-1+\sqrt{2}}{2}
.\end{array}
Hence, $
x=\left\{ \dfrac{-1-\sqrt{2}}{2},\dfrac{-1+\sqrt{2}}{2} \right\}
.$
