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If x-22-5|x-2...

Question

If $(x - 2)^{2} - 5 | x - 2 | + 6 = 0$ , then $x \in$

{0, 4}

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{- 1, 5}

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{- 1, 0, 4, 5}

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{2, 3}

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Solution

The correct option is C ${- 1, 0, 4, 5}$
$(x - 2)^{2} - 5 | x - 2 | + 6 = 0 \Rightarrow | x - 2 |^{2} - 5 | x - 2 | + 6 = 0 (∵ x^{2} = | x |^{2})$

Let $| x - 2 | = t$
$\Rightarrow t^{2} - 5 t + 6 = 0$
$\Rightarrow t^{2} - 2 t - 3 t + 6 = 0$
$\Rightarrow (t - 2) (t - 3) = 0$
$\Rightarrow t = 2, 3$
$\Rightarrow | x - 2 | = 2 or | x - 2 | = 3$
$\Rightarrow x - 2 = \pm 2 or x - 2 = \pm 3$
$\Rightarrow x = 0, 4 or x = - 1, 5$
$∴ x \in {- 1, 0, 4, 5}$

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