Determine real roots of AR(1) process

Question

Can someone please help me understand why the real roots of the polynomial

$\phi$(B) = 1 - $\frac{1}{3}B$ - $\frac{1}{2}B^2$ are

$ = \frac{-2 \pm\sqrt{76}}{6}$

Thank you in advance.

Answer 1

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form:\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$ $$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

in your case, when

$$1 - \frac{1}{3}B - \frac{1}{2}B^2 =0$$

Multiply both sides by $6$ and re-arrange: $$-3B^2-2B+6=0$$

Comparing to standard formula, we get:

$$a=-3, b=-2, C=6$$

Substitute into the standard formula, you get:

$$B_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\left(-3\right)6}}{2\left(-3\right)}$$

You could simplify to get your answer.