Đáp án: $\,{y^2} - \dfrac{{11}}{2}y + \dfrac{{15}}{2} = 0$

Giải thích các bước giải:

Gọi pt chứa 2 nghiệm cần tìm là 

$\begin{array}{l}
{y^2} - S.y + P = 0\\
 \Leftrightarrow \left\{ \begin{array}{l}
S = {y_1} + {y_2}\\
P = {y_1}.{y_2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
S = \left| {{x_1}} \right| + 2 + \left| {{x_2}} \right| + 2 = \left| {{x_1}} \right| + \left| {{x_2}} \right| + 4\\
P = \left( {\left| {{x_1}} \right| + 2} \right)\left( {\left| {{x_2}} \right| + 2} \right) = \left| {{x_1}{x_2}} \right| + 2\left( {\left| {{x_1}} \right| + \left| {{x_2}} \right|} \right) + 4
\end{array} \right.\\
Do:2{x^2} - 3x - 1 = 0\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{3}{2}\\
{x_1}{x_2} =  - \dfrac{1}{2}
\end{array} \right.\\
Xet:\left| {{x_1}} \right| + \left| {{x_2}} \right|\\
 = \sqrt {{{\left( {\left| {{x_1}} \right| + \left| {{x_2}} \right|} \right)}^2}} \\
 = \sqrt {x_1^2 + x_2^2 + 2\left| {{x_1}{x_2}} \right|} \\
 = \sqrt {{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2} - 2\left| { - \dfrac{1}{2}} \right|} \\
 = \sqrt {{{\left( {\dfrac{3}{2}} \right)}^2} - 2.\dfrac{{ - 1}}{2} - 2.\dfrac{1}{2}} \\
 = \dfrac{3}{2}\\
 \Leftrightarrow \left\{ \begin{array}{l}
S = \dfrac{3}{2} + 4 = \dfrac{{11}}{2}\\
P = \left| {\dfrac{{ - 1}}{2}} \right| + 2.\dfrac{3}{2} + 4 = \dfrac{{15}}{2}
\end{array} \right.\\
Vậy\,{y^2} - \dfrac{{11}}{2}y + \dfrac{{15}}{2} = 0
\end{array}$