Have you learnt Newton sum?

Algebra Level 4

A polynomial $5x^2-6x-2=0$ has roots $\alpha$ and $\beta$ . Then, find $(\alpha^5+\alpha^{10}+\beta^{10}+\beta^5).$

2 solutions

Jesse Nieminen
Jun 9, 2016

Vieta's Formula tells that $\large\alpha + \beta = \frac{6}{5} = e_1$ and $\large\alpha\beta = -\frac{2}{5} = e_2$ .

Now, $\large P_n = \alpha^n + \beta^n$ can be computed using Newton's Identities for any positive integer $\large n$ .

\large P_1 =

\large e_1 = \frac{6}{5}

\large P_2 =

\large e_1 P_1 - e_2\times2 = \frac{56}{25}

\large P_3 =

\large e_1P_2 - e_2P_1 = \frac{396}{25}

\large P_4 =

\large e_1P_3 - e_2P_2 = \frac{2936}{25}

\large P_5 =

\large e_1P_4 - e_2P_5 = \frac{21576}{5^5}

\large P_{10} =

\large{P_5}^2 - {e_2}^5\times2 = \frac{465723776}{5^{10}}

Using the calculations above we find that $\large \alpha^5+\alpha^{10}+\beta^{10}+\beta^5 = \frac{533148776}{5^{10}} = \boxed{54.5944346624}$ .

Good approach. Glad you decided against calculating $P_6, P_7, P_8, P_9$ .

How did you directly calculated $P_{10}$ ?? Is there any formula or shortcut to such manupulation ?? Thank you .

Chirayu Bhardwaj - 4 years, 11 months ago

Yosua Sibuea
Dec 1, 2015