Question

7,8.Восьмое желательно быстрее.Тема виета класс 8

Answer 1

$7)\ \ 3x^2-bx+14=0 \ \ \ ,\ \ \ x_1=5\\\\teorema\ Vieta:\ \left\{\begin{array}{l}\ x_1+x_2=\dfrac{b}{3}\\x_1\cdot x_2=\dfrac{14}{3}\end{array}\right\ \ \left\{\begin{array}{l}\ 5+x_2=\dfrac{b}{3}\\5\cdot x_2=\dfrac{14}{3}\end{array}\right\ \ \left\{\begin{array}{l}\ x_2=\dfrac{14}{15}\\5+\dfrac{14}{15}=\dfrac{b}{3}\end{array}\right$

$\left\{\begin{array}{l}\ x_2=\dfrac{14}{15\\\dfrac{b}{3}}=\dfrac{89}{15}\end{array}\right\ \ \ \left\{\begin{array}{l}\ x_2=\dfrac{14}{15}\\b=\dfrac{89}{5}\end{array}\right\\\\\\Otvet:\ \ x_2=\dfrac{14}{15}\ ,\ \ b=\dfrac{89}{5}\ .$

$8)\ \ 8x^2+20x-c=0 \ \ \ ,\ \ \ x_1=2,5\\\\teorema\ Vieta:\ \left\{\begin{array}{l}x_1+x_2=-\dfrac{20}{8}\\\ x_1\cdot x_2=-\dfrac{c}{8}\end{array}\right\ \ \left\{\begin{array}{l}\ 2,5+x_2=-\dfrac{5}{2}\\\ 2,5\cdot x_2=-\dfrac{c}{8}\end{array}\right\ \ \left\{\begin{array}{l}\ x_2=-5\\-2,5\cdot 5=-\dfrac{c}{8}\end{array}\right\\\\\\\left\{\begin{array}{l}x_2=-5\\c=100\end{array}\right\ \ \ \Rightarrow \ \ \ 8x^2+20x-100=0$

$Otvet:\ \ x_2=-5\ ,\ c=100\ .$