If $x^{2} - 5 x sgn (x^{2} - 4) + 6 = 0,$ then number of values of $x$ satisfying it is

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Solution

Case $1 : x^{2} - 4 < 0 \Rightarrow - 2 < x < 2$
$\Rightarrow x^{2} - 5 x (- 1) + 6 = 0 \Rightarrow x^{2} + 5 x + 6 = 0 \Rightarrow x = - 2, - 3$
$∴$ No solution in $- 2 < x < 2$

Case $2 : x^{2} - 4 = 0 \Rightarrow x = - 2, 2$
$\Rightarrow x^{2} - 5 (0) + 6 = 0 \Rightarrow x^{2} + 6 = 0$
$∴$ No solution in $x = \pm 2$

Case $3 : x^{2} - 4 > 0$
$\Rightarrow x \in (- \infty, - 2) \cup (2, \infty)$
$\Rightarrow x^{2} - 5 x (1) + 6 = 0 \Rightarrow x^{2} - 5 x + 6 = 0 \Rightarrow (x - 2) (x - 3) = 0 \Rightarrow x = 3$

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