∣x^2 −4x∣ +∣x∣ ≥ 3


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Question Number 139695 by john_santu last updated on 30/Apr/21

$∣ x^{2} - 4 x ∣ + ∣ x ∣ ⩾ 3$
	Answered by MJS_new last updated on 30/Apr/21

	$∣ x (x - 4) ∣ + ∣ x ∣= 3$ $squaring$ $x^{2} {(1 + ∣ x - 4 ∣)}^{2} = 9$ $2 x^{2} ∣ x - 4 ∣= - x^{4} + 8 x^{3} - 17 x^{2} + 9$ $squaring & transforming$ $x^{8} - 16 x^{7} + 94 x^{6} - 240 x^{5} + 207 x^{4} + 144 x^{3} - 306 x^{2} + 81 = 0$ $x = t + 2$ $t^{8} - 18 t^{6} - 8 t^{5} + 87 t^{4} + 72 t^{3} - 74 t^{2} - 168 t - 135 = 0$ $(t^{2} - t - 9) (t^{2} - t - 3) (t^{2} + t - 5) (t^{2} + t + 1) = 0$ $this is easy to solve but only 2 solutions fit$ $the original equation$ $x_{1} = \frac{5 - \sqrt{37}}{2} \land x_{2} = \frac{5 - \sqrt{13}}{2}$ $\Rightarrow$ $x ⩽ \frac{5 - \sqrt{37}}{2} \lor x ⩾ \frac{5 - \sqrt{13}}{2}$
	Answered by mr W last updated on 30/Apr/21

	$f o r x ⩽ 0 :$ $x^{2} - 4 x - x ⩾ 3$ $x^{2} - 5 x - 3 ⩾ 0$ $x ⩽ \frac{5 - \sqrt{37}}{2}$ $f o r 0 ⩽ x ⩽ 4 :$ $- x^{2} + 4 x + x ⩾ 3$ $x^{2} - 5 x + 3 ⩽ 0$ $\frac{5 - \sqrt{13}}{2} ⩽ x ⩽ \frac{5 + \sqrt{13}}{2}$ $\Rightarrow \frac{5 - \sqrt{13}}{2} ⩽ x ⩽ 4$ $f o r x ⩾ 4 :$ $x^{2} - 4 x + x ⩾ 3$ $x^{2} - 3 x - 3 ⩾ 0$ $x ⩾ \frac{3 + \sqrt{21}}{2}$ $\Rightarrow x ⩾ 4$ $s u m m a r y :$ $x ⩽ \frac{5 - \sqrt{37}}{2} o r x ⩾ \frac{5 - \sqrt{13}}{2}$
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