Straight LineHardBloom L3
Question
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 equation $y^{2} - 3y + 2 = 0$, then the possible vertices of the square is/are:
Options
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.$( - 2,1),( - 1, - 1),( - 1,2),( - 2,2)$
Solution
Sol. [Key Concept: Roots of $x^{2 - 3}|x| + 2 = 0$ give $x = \pm 1, \pm 2$; with $yin\{ 1,2\}$ the unit squares are fixed.]
Possible squares have vertices
$$( - 2,1),( - 2,2),( - 1,1),( - 1,2)$$
or $(1,1),(1,2),(2,1),(2,2)$
✔ (a) vertices $( - 2,1),( - 2,2),( - 1,1),( - 1,2)$ form a unit square.
✔ (b) vertices $(1,1),(1,2),(2,1),(2,2)$ form a unit square.
✘ (c) not a unit square from these roots.
✘ (d) not a unit square from these roots.
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