Question

Exercitiul 21 ..
Multumesc anticipat :))

Answer 1

$x^2+3ax-a+3 = 0$

$\\a)\,\,\,x_1 = 2x_2\\ \\ x_1+x_2 = -3a \Rightarrow 2x_2+x_2 = -3a \Rightarrow 3x_2 = -3a \Rightarrow x_2 = -a\\ x_1x_2 = -a+3 \Rightarrow x_2\cdot(-a) = -a+3 \Rightarrow x_1 = \dfrac{a-3}{a}\\ \\ \dfrac{a-3}{a} = 2\cdot (-a) \Rightarrow a-3 = -2a^2 \Rightarrow 2a^2+a-3 = 0\\ \\ \Delta = 1+24 = 25 \Rightarrow a_{1,2} = \dfrac{-1\pm 5}{4}\Rightarrow \left|\begin{array}{l} a_{1} =\frac{-6}{4} = -\frac{3}{2}\\\\a_{2} = \frac{4}{4} = 1\end{array}\right.\\ \Rightarrow \boxed{a\in \left\{-\dfrac{3}{2}, 1\right\}}$

$\\b)\,\,\,x_1^2+x_2^2 \leq 5 \Rightarrow (x_1+x_2)^2-2x_1x_2 \leq 5 \Rightarrow\\ \\ \Rightarrow (-3a)^2-2(-a+3)\leq 5 \Rightarrow 9a^2+2a-6\leq 5 \Rightarrow \\ \\ \Rightarrow 9a^2+2a-11 \leq 0\\ \\ \Delta = 4+396 = 400 \Rightarrow a_{1,2} = \dfrac{-2\pm 20}{18}\\ \\ \Rightarrow \boxed{a\in \left[-\dfrac{11}{9},1\right]}$