Find the solution of inequality : x^2 + ∣x∣


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Question Number 63481 by naka3546 last updated on 04/Jul/19

$F i n d t h e s o l u t i o n o f i n e q u a l i t y :$ $x^{2} + ∣ x ∣ > 6$
Commented bymathmax by abdo last updated on 04/Jul/19
$(ine) ⇒∣x∣^2 +∣x∣−6>0 ⇒t^2 +t−6>0 Δ=1−4(−6)=25 ⇒ t_1 =((−1+5)/2) =2 and t_2 =((−1−5)/2) =−3 ⇒ ∣x∣^2 +∣x∣−6 =(∣x∣−2)(∣x∣+3) so (ine) ⇒{∣x∣−2}{∣x∣+3}>0 ⇒ ∣x∣−2 >0 (because ∣x∣ +3>0 ⇒ x>2 or x<−2 ⇒ x ∈]−∞,−2[∪]2,+∞[$
$(i n e) \Rightarrow∣ x ∣^{2} + ∣ x ∣ - 6 > 0 \Rightarrow t^{2} + t - 6 > 0$ $Δ = 1 - 4 (- 6) = 25 \Rightarrow t_{1} = \frac{- 1 + 5}{2} = 2 a n d t_{2} = \frac{- 1 - 5}{2} = - 3 \Rightarrow$ $∣ x ∣^{2} + ∣ x ∣ - 6 = (∣ x ∣ - 2) (∣ x ∣ + 3) s o (i n e) \Rightarrow {∣ x ∣ - 2} {∣ x ∣ + 3} > 0 \Rightarrow$ $∣ x ∣ - 2 > 0 (b e c a u s e ∣ x ∣ + 3 > 0 \Rightarrow x > 2 o r x < - 2 \Rightarrow$ $x \in] - \infty, - 2 [\cup] 2, + \infty [$
	Answered by MJS last updated on 04/Jul/19

	$x > 0$ $x^{2} + x > 6$ $x^{2} + x - 6 = 0 \Rightarrow x = - 3 \lor x = 2$ $\Rightarrow x < - 3 \lor x > 2 but x > 0$ $\Rightarrow x > 2$ $x < 0$ $x^{2} - x > 6$ $x^{2} - x - 6 = 0 \Rightarrow x = - 2 \lor x = 3$ $\Rightarrow x < - 2 \lor x > 3 but x < 0$ $\Rightarrow x < - 2$ $x^{2} + ∣ x ∣> 6 \Rightarrow x < - 2 \lor 2 < x \Leftrightarrow x \in] - \infty; - 2 [\cup] 2; + \infty [$
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