Answer
$p=\left\{ \dfrac{-1-\sqrt{7}}{6},\dfrac{-1+\sqrt{7}}{6}
\right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
p^2+\dfrac{p}{3}=\dfrac{1}{6}
,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula.
$\bf{\text{Solution Details:}}$
Using the properties of equality, in the form $ax^2+bx+c=0,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
p^2+\dfrac{p}{3}-\dfrac{1}{6}=0
\\\\
6\left(p^2+\dfrac{p}{3}-\dfrac{1}{6}\right)=6(0)
\\\\
6p^2+2p-1=0
.\end{array}
The quadratic equation above has $a=
6
, b=
2
, c=
-1
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
p=\dfrac{-2\pm\sqrt{2^2-4(6)(-1)}}{2(6)}
\\\\
p=\dfrac{-2\pm\sqrt{4+24}}{12}
\\\\
p=\dfrac{-2\pm\sqrt{28}}{12}
.\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}
p=\dfrac{-2\pm\sqrt{4\cdot7}}{12}
\\\\
p=\dfrac{-2\pm\sqrt{(2)^2\cdot7}}{12}
\\\\
p=\dfrac{-2\pm2\sqrt{7}}{12}
.\end{array}
Cancelling the common factors from all the terms results to
\begin{array}{l}\require{cancel}
p=\dfrac{\cancel2(-1)\pm\cancel2(1)\sqrt{7}}{\cancel2(6)}
\\\\
p=\dfrac{-1\pm\sqrt{7}}{6}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
p=\dfrac{-1-\sqrt{7}}{6}
\\\\\text{OR}\\\\
p=\dfrac{-1+\sqrt{7}}{6}
.\end{array}
Hence, $
p=\left\{ \dfrac{-1-\sqrt{7}}{6},\dfrac{-1+\sqrt{7}}{6}
\right\}
.$
