Question

Considere as raízes da equação
$2 {x }^{2} - 3x + k = 0$
Determine K para que se tenha
${a}^{3} + {b}^{3} = \frac{ - 243}{8}$

Answer 1

$2x^2-3x+k=0\ \to\ a=2,\ b=-3,\ c=k\\\\ S_1=\alpha+\beta=-\dfrac{b}{a}=\dfrac{3}{2}\\\\ (\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta\ \to\ \bigg(\dfrac{3}{2}\bigg)^2=\alpha^2+\beta^2+2\bigg(\dfrac{c}{a}\bigg)\ \to\\\\ \to\ \alpha^2+\beta^2=\dfrac{9}{4}-2\bigg(\dfrac{k}{2}\bigg)\ \to\ S_2=\dfrac{9}{4}-k\\\\ S_3=\alpha^3+\beta^3=-\dfrac{243}{8}$

$aS_{n+2}+bS_{n+1}+cS_n=0\ \to\ 2S_3-3S_2+kS_1=0\ \to\\\\ \to\ 2\bigg(\dfrac{-243}{8}\bigg)-3\bigg(\dfrac{9}{4}-k\bigg)+k\bigg(\dfrac{3}{2}\bigg)=0\ \to\\\\ \to\ -\dfrac{243}{4}-\dfrac{27}{4}+3k+\dfrac{3k}{2}=0\ \to\ \dfrac{6k}{2}+\dfrac{3k}{2}=\dfrac{270}{4}\ \to\\\\ \to\ 9k=\dfrac{270}{2}\ \to\ k=\dfrac{135}{9}\ \to\ \boxed{k=15}$