az^2 + z + 1/a = 0 có | z1| + | z2| = 2 . tim a

Đáp án: $a =  \pm \frac{1}{2}$

Giải thích các bước giải:

$\begin{array}{l}
a \ne 0\\
Theo\,Viet:\left\{ \begin{array}{l}
{z_1} + {z_2} = \frac{{ - 1}}{a}\\
{z_1}{z_2} = \frac{1}{{{a^2}}}
\end{array} \right.\\
\left| {{z_1}} \right| + \left| {{z_2}} \right| = 2\\
 \Rightarrow z_1^2 + 2\left| {{z_1}{z_2}} \right| + z_2^2 = 4\\
 \Rightarrow {\left( {{z_1} + {z_2}} \right)^2} - 2{z_1}{z_2} + 2.{z_1}{z_2} = 4\\
\left( {do:{z_1}{z_2} = \frac{1}{{{a^2}}} > 0} \right)\\
 \Rightarrow {\left( { - \frac{1}{a}} \right)^2} = 4\\
 \Rightarrow {a^2} = \frac{1}{4}\\
 \Rightarrow a =  \pm \frac{1}{2}
\end{array}$