The value of $x : ∣ ∣ x^{2} + 2 x - 8 ∣ ∣ + x - 2 = 0$ is ?

- 5

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- 2

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2

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- 3

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Solution

The correct option is C $2$
Case $1 : x > 0$
$∣ ∣ x^{2} + 2 x - 8 ∣ ∣ + x - 2 = 0$
$\Rightarrow x^{2} + 2 x - 8 + x - 2 = 0$
$\Rightarrow x^{2} + 3 x - 10 = 0$
$\Rightarrow x^{2} + 5 x - 2 x - 10 = 0$
$\Rightarrow x (x + 5) - 2 (x + 5) = 0$
$\Rightarrow (x + 5) (x - 2) = 0$
$\Rightarrow x = {- 5, 2}$
Case $2 : x < 0$
$∣ ∣ x^{2} + 2 x - 8 ∣ ∣ + x - 2 = 0$
$\Rightarrow - x^{2} - 2 x + 8 + x - 2 = 0$
$\Rightarrow - x^{2} - x + 6 = 0$
$\Rightarrow x^{2} + x - 6 = 0$
$\Rightarrow x^{2} + 3 x - 2 x - 6 = 0$
$\Rightarrow x (x + 3) - 2 (x + 3) = 0$
$\Rightarrow (x + 3) (x - 2) = 0$
$\Rightarrow x = {- 3, 2}$
$x = {- 5, 2} \cap {- 3, 2} = {2}$

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