# Determine real roots of AR(1) process [closed]

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.

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

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.

• Thank so so much! Can you please also explain where the multiplier "6" comes from? – Nadine M. Sep 4 '19 at 13:18
• You don't have to multiply by $6$. However I did that to simplify the equation and hence the formula for the root. In this case, 6 was chosen because you have denominators of 3 and 2 and I wanted no fractions for ease of calculations. A choice of (-6) would even be better but confuses signs. – NoChance Sep 4 '19 at 13:20
• Does make sense, thanks again! – Nadine M. Sep 4 '19 at 13:26
• You are always welcome :) – NoChance Sep 4 '19 at 13:27