Question

7 задание,8 класс тема виета

Answer 1

Ответ:

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

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