Question

9 и 10.Если можете другие тоже то желательно.Объясните плз

Answer 1

$9)\ \ x^2-15x+q=0 \ \ \ ,\ \ \ x_1-x_2=3\ \ \Rightarrow \ \ x_1=x_2+3\\\\teorema\ Vieta:\ \left\{\begin{array}{l}\ x_1+x_2=15\\x_1\cdot x_2=q\end{array}\right\ \ \left\{\begin{array}{l}(x_2+3)+x_2=15\\x_1\cdot x_2=q\end{array}\right\ \ \left\{\begin{array}{l}2x_2=12\\x_1\cdot x_2=q\end{array}\right$

$\left\{\begin{array}{l}\ x_2=6\ ,\ x_1=9\\6\cdot 9=q\end{array}\right\ \ \ \left\{\begin{array}{l}\ x_2=6\\q=54\end{array}\right\\\\\\Otvet:\ \ q=54\ .$

$10)\ \ 4x^2+2x+c=0 \ \ \ ,\ \ \ x_1-x_2=3,5\ \ \Rightarrow \ \ x_1=x_2+3,5\\\\teorema\ Vieta:\ \left\{\begin{array}{l}\ x_1+x_2=-\dfrac{2}{4}\\x_1\cdot x_2=\dfrac{c}{4}\end{array}\right\ \ \left\{\begin{array}{l}(x_2+3,5)+x_2=-0,5\\x_1\cdot x_2=\dfrac{c}{4}\end{array}\right\ \ \left\{\begin{array}{l}2x_2=-4\\x_1\cdot x_2=\dfrac{c}{4}\end{array}\right$

$\left\{\begin{array}{l}\ x_2=-2\ ,\ x_1=1,5\\-2\cdot 1,5=c\end{array}\right\ \ \ \left\{\begin{array}{l}\ x_2=-2\\c=-3\end{array}\right\\\\\\Otvet:\ \ c=-3\ .$