The x-coordinates of the vertices of a square | vedclass

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(D) For the $x$-coordinates: $x^2 - 3|x| + 2 = 0$. Let $|x| = t$,then $t^2 - 3t + 2 = 0$,which gives $(t-1)(t-2) = 0$. So,$|x| = 1$ or $|x| = 2$,which

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Both $(A)$ and $(B)$

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(D) For the $x$-coordinates: $x^2 - 3|x| + 2 = 0$. Let $|x| = t$,then $t^2 - 3t + 2 = 0$,which gives $(t-1)(t-2) = 0$. So,$|x| = 1$ or $|x| = 2$,which implies $x \in \{1, -1, 2, -2\}$. For the $y$-coordinates: $y^2 - 3y + 2 = 0$. This gives $(y-1)(y-2) = 0$,so $y \in \{1, 2\}$. The square has unit area,meaning the side length is $1$. Possible sets of vertices forming a square of side $1$ are: $S_1 = \{(1, 1), (2, 1), (2, 2), (1, 2)\}$ $S_2 = \{(-1, 1), (-2, 1), (-2, 2), (-1, 2)\}$ Both sets represent squares of unit area. Thus,both $(A)$ and $(B)$ are correct.

The $x$-coordinates of the vertices of a square of unit area are the roots of the equation $x^2 - 3|x| + 2 = 0$ 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 are:

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