Answer: The correct option is (B) [tex]\dfrac{-1-i\sqrt2}{3}.[/tex]
Step-by-step explanation: Given that Jonathan must determine the solutions of the quadratic equation :
[tex]3x^2+2x+1=0.[/tex]
We are to select the correct solution to the above quadratic equation.
The solution of a quadratic equation of the form [tex]ax^2+bx+c=0,~a\neq 0[/tex] is given by
[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.[/tex]
In the given equation, we have
[tex]a=3,~~b=2,~~c=1.[/tex]
Therefore, the solution of the equation is given by
[tex]x\\\\\\=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\\=\dfrac{-2\pm\sqrt{2^2-4\times3\times1}}{2\times3}\\\\\\=\dfrac{-2\pm\sqrt{4-12}}{6}\\\\\\=\dfrac{-2\pm\sqrt{-8}}{6}\\\\\\=\dfrac{-2\pm\sqrt{8i^2}}{6}~~~~~~~~~~~~~~~~~~\textup{since }i^2=1\\\\\\=\dfrac{-2\pm2i\sqrt2}{6}\\\\\\=\dfrac{-1\pm i\sqrt2}{3}.[/tex]
So, the solutions of the given equation are
[tex]x=\dfrac{-1+i\sqrt2}{3},~~~~~\dfrac{-1-i\sqrt2}{3}.[/tex]
Out of the given options, a solution of the equation is [tex]\dfrac{-1-i\sqrt2}{3.}[/tex]
Option (B) is CORRECT.
