$| x^{2} + 2 x - 8 | + x - 2 = 0$

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Solution

$(x^{2} + 2 x - 8) + (x - 2) = 0$
$(x^{2} + 4 x - 2 x - 8) + (x - 2) = 0$
$[x (x + 4) - 2 (x + 4)] + (x - 2) = 0$
$[(x - 2) (x + 4)] + (x - 2) = 0$
$[x - 2] [1 + (x + 4)] = 0$
$x = 2, x = - 5$

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Similar questions

x : ∣ ∣ x^{2} + 2 x - 8 ∣ ∣ + x - 2 = 0

x = - 2

x^{2} - 2 x + 8 = 0

x^{2} + 2 x - 8 = 0

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$(x) {lim}_{x \to 0^{+}} (2 - c o t x)$

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{x \in R ∣ ∣ x^{2} - 2 x - 8 \leq 0}

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