The $x -$ co-ordinates of the vertices of a square of unit area are the roots of the equation $x^2 - 3 |x| + 2 = 0$ and the $y -$ co-ordinates of the vertices are the roots of the equation $y^2 - 3y + 2 = 0$ then the possible vertices of the square is/are :
The $x -$ co-ordinates of the vertices of a square of unit area are the roots of the equation $x^2 - 3 |x| + 2 = 0$ and the $y -$ co-ordinates of the vertices are the roots of the equation $y^2 - 3y + 2 = 0$ then the possible vertices of the square is/are :
- A
$(1, 1), (2, 1), (2, 2), (1, 2)$
- B
$(- 1, 1), (- 2, 1), (- 2, 2), (- 1, 2)$
- C
$(2, 1), (1, - 1), (1, 2), (2, 2)$
- D
$(A)$ or $(B)$ both
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