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.

  • $\begingroup$ Thank so so much! Can you please also explain where the multiplier "6" comes from? $\endgroup$ – Nadine M. Sep 4 '19 at 13:18
  • $\begingroup$ 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. $\endgroup$ – NoChance Sep 4 '19 at 13:20
  • $\begingroup$ Does make sense, thanks again! $\endgroup$ – Nadine M. Sep 4 '19 at 13:26
  • $\begingroup$ You are always welcome :) $\endgroup$ – NoChance Sep 4 '19 at 13:27

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