Question

{х2-4х-12<0\х2-6х-7<0​

Answer 1

Ответ:

$\left\{\begin{array}{l}x^2-4x-12<0\\x^2-6x-7<0\end{array}\right\ \ \left\{\begin{array}{l}(x+2)(x-6)<0\\(x+1)(x-7)<0\end{array}\right\ \ \left\{\begin{array}{l}x\in (-2;\, 6\, )\\x\in (-1\, ;\, 7\, )\end{array}\right\ \ \Rightarrow \\\\\\\boxed{\ x\in (-1\, ;\, 6\, )\ }\\\\\\\star \ x^2-4x-12=0\ \ ,\ \ x_1=-2\ ,\ x_2=6\ \ (teorema\ Vieta)\ \star \\\\\star \ x^2-6x-7=0\ \ ,\ \ x_1=-1\ ,\ x_2=7\ \ (teorema\ Vieta)\ \star$