Christmas Questions: No. 2

Algebra Level pending

Let $x_{1},x_{2},x_{3},x_{4},x_{5}$ be the solutions for the polynomial

$24x^{5}+178x^{4}+255x^{3}-417x^{2}-6x-1781=0$

If $\displaystyle \sum_{i=1}^5 x_i + \prod_{i=1}^5 x_i$ can be expressed as $\dfrac AB$ , calculate $A+B$ .

1 solution

Oliver Papillo
Dec 27, 2017

By Vieta's Formulae : $\displaystyle \sum_{i=1}^5 x_i = \frac{-178}{24}$

$\displaystyle \prod_{i=1}^5 x_i = \frac{-(-1781)}{24} = \frac{1781}{24}$

So $\displaystyle \sum_{i=1}^5 x_i + \prod_{i=1}^5 x_i = \frac{1781 - 178}{24} = \frac{1603}{24}$

So $A = 1603$ , $B = 24$ , and $A+B = 1627$

Brilliant solution! Thank you for posting this.

Kok Hao - 3 years, 5 months ago