Question

Задание в прикрепленном фото

Answer 1

Ответ:

$1)\ \ x^2+ax-21=0\ \ ,\ \ x_1=-3\\\\teorema\ Vieta:\ \ \left\{\begin{array}{l}-3+x_2=-a\\-3\cdot x_2=-21\end{array}\right\ \ \left\{\begin{array}{l}x_2=7\\-3+7=-a\end{array}\right\ \ \left\{\begin{array}{l}x_2=7\\a=-4\end{array}\right\\\\\\Otvet:\ \ a=-4\ \ ,\ \ x_1=-3\ ,\ \ x_2=7\ .$

$2)\ \ a^2x^2-4ax-5=0\ \ ,\ \ x_1=-\dfrac{1}{4}\ \ \ \ (a\ne 0)\\\\teorema\ Vieta:\ \ \left\{\begin{array}{l}-\dfrac{1}{4}+x_2=\dfrac{4a}{a^2}\\\ \, -\dfrac{1}{4}\cdot x_2=-\dfrac{5}{a^2}\end{array}\right\ \ \left\{\begin{array}{l}x_2=\dfrac{4}{a}+\dfrac{1}{4}\\\ \, x_2=\dfrac{20}{a^2}\end{array}\right\ \ \left\{\begin{array}{l}\dfrac{20}{a^2}=\dfrac{4}{a}+\dfrac{1}{4}\\\ \, x_2=\dfrac{20}{a^2}\end{array}\right$

$\displaystyle \frac{20}{a^2}=\frac{4}{a}+\frac{1}{4}\ \ \ ,\ \ \ \frac{20}{a^2}-\frac{4}{a}-\frac{1}{4}=0\ \ ,\ \ \ \frac{80-16a-a^2}{4a^2}=0\ \ ,\\\\\\a^2+16a-80=0\ \ \ \Rightarrow \ \ \ a_1=-20\ ,\ \ a_2=4\ \ (teorema\ Vieta)\\\\Otvet:\ \ a_1=-20\ ,\ a_2=4\ .$