The \[x\]-coordinates of the vertices of a square of unit area are the roots of the equation \[{x^2} - 3\mid x\mid + 2 = 0\] and the \[y\]-coordinates of the vertices are the roots of the equation \[{y^2} - 3y + 2 = 0\] . Then the possible vertices of the square is/are
This question has multiple correct options
A) \[\left( {1,1} \right)\] , \[\left( {2,1} \right)\] , \[\left( {2,2} \right)\] , \[\left( {1,2} \right)\]
B) \[\left( { - 1,1} \right)\] , \[\left( { - 2,1} \right)\] , \[\left( { - 2,2} \right)\] , \[\left( { - 1,2} \right)\]
C) \[\left( {2,1} \right)\] , \[\left( {1, - 1} \right)\] , \[\left( {1,2} \right)\] , \[\left( {2,2} \right)\]
D) \[\left( { - 2,1} \right)\] , \[\left( { - 1, - 1} \right)\] , \[\left( { - 1,2} \right)\] , \[\left( { - 2,2} \right)\]

Hint:
Here we need to find the coordinates of the square. We will first find the roots of the first equation to get the \[x\] coordinates of the vertices of the square. Then we will find the roots of the second equation to get the \[y\] coordinates of vertices of the square. From there, we will get all the possible vertices of the square.

Complete step by step solution:
It is given that \[x\] coordinates of the vertices of the square are the roots of the equation \[{x^2} - 3\mid x\mid + 2 = 0\]
Now, we will find the roots of this quadratic equation.
\[ \Rightarrow {x^2} - 3\mid x\mid + 2 = 0\]
Now, we will factorize this quadratic equation by splitting the middle terms.
\[ \Rightarrow {x^2} - 2\mid x\mid - \left| x \right| + 2 = 0\]
On further simplification, we get
\[ \Rightarrow \left( {\left| x \right| - 2} \right)\left( {\left| x \right| - 2} \right) = 0\]
Using zero product property, we get
\[ \Rightarrow \left( {\left| x \right| - 1} \right) = 0\] or \[\left( {\left| x \right| - 2} \right) = 0\]
Simplifying the equations, we get
\[ \Rightarrow \left| x \right| = 1\] or \[\left| x \right| = 2\]
We know that by removing the modulus, we will give two values.
Therefore, we get
\[x = - 1,1\] and \[x = - 2,2\]
Hence, the total possible values of \[x\] are \[ - 2\], \[ - 1\], 1 and 2.
Now, we will find the roots of this quadratic equation.
\[{y^2} - 3y + 2 = 0\]
Now, we will factorize this quadratic equation by splitting the middle terms. Therefore, we get
\[{y^2} - 2y - y + 2 = 0\]
On further simplification, we get
\[ \Rightarrow \left( {y - 1} \right)\left( {y - 2} \right) = 0\]
Using the zero product property, we get
\[ \Rightarrow \left( {y - 1} \right) = 0\] or \[\left( {y - 2} \right) = 0\]
Simplifying the equation, we get
\[ \Rightarrow y = 1\] and \[y = 2\]
Hence, the total possible values of \[y\] are 1 and 2.
It is given that the given square is of unit area. So the coordinates of the vertices of the square are
\[\left( {1,1} \right)\] , \[\left( {2,1} \right)\] , \[\left( {2,2} \right)\] , \[\left( {1,2} \right)\]
And also the coordinates of the vertices are:
\[\left( { - 1,1} \right)\] , \[\left( { - 2,1} \right)\] , \[\left( { - 2,2} \right)\] , \[\left( { - 1,2} \right)\]

Hence, the correct options are option A and option B.

Note:
Here, we have factorize the quadratic equation to find the roots of the equation. Roots of the quadratic equation are defined as the values of the variable which when put in the quadratic equation satisfy the equation. We need to keep in mind that the number of roots of the equation is equal to highest power of the equation. But here we got 4 values of \[x\] from the equation \[{x^2} - 3\mid x\mid + 2 = 0\] and not 2 because here we have modulus sign in the equation.