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Question Number 139695 by john_santu last updated on 30/Apr/21
∣x^2 −4x∣ +∣x∣ ≥ 3
$$\:\mid{x}^{\mathrm{2}} −\mathrm{4}{x}\mid\:+\mid{x}\mid\:\geqslant\:\mathrm{3}\: \\ $$
Answered by MJS_new last updated on 30/Apr/21
∣x(x−4)∣+∣x∣=3 squaring x^2 (1+∣x−4∣)^2 =9 2x^2 ∣x−4∣=−x^4 +8x^3 −17x^2 +9 squaring & transforming x^8 −16x^7 +94x^6 −240x^5 +207x^4 +144x^3 −306x^2 +81=0 x=t+2 t^8 −18t^6 −8t^5 +87t^4 +72t^3 −74t^2 −168t−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 =((5−(√(37)))/2)∧x_2 =((5−(√(13)))/2) ⇒ x≤((5−(√(37)))/2)∨x≥((5−(√(13)))/2)
$$\mid{x}\left({x}−\mathrm{4}\right)\mid+\mid{x}\mid=\mathrm{3} \\ $$$$\mathrm{squaring} \\ $$$${x}^{\mathrm{2}} \left(\mathrm{1}+\mid{x}−\mathrm{4}\mid\right)^{\mathrm{2}} =\mathrm{9} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} \mid{x}−\mathrm{4}\mid=−{x}^{\mathrm{4}} +\mathrm{8}{x}^{\mathrm{3}} −\mathrm{17}{x}^{\mathrm{2}} +\mathrm{9} \\ $$$$\mathrm{squaring}\:\&\:\mathrm{transforming} \\ $$$${x}^{\mathrm{8}} −\mathrm{16}{x}^{\mathrm{7}} +\mathrm{94}{x}^{\mathrm{6}} −\mathrm{240}{x}^{\mathrm{5}} +\mathrm{207}{x}^{\mathrm{4}} +\mathrm{144}{x}^{\mathrm{3}} −\mathrm{306}{x}^{\mathrm{2}} +\mathrm{81}=\mathrm{0} \\ $$$${x}={t}+\mathrm{2} \\ $$$${t}^{\mathrm{8}} −\mathrm{18}{t}^{\mathrm{6}} −\mathrm{8}{t}^{\mathrm{5}} +\mathrm{87}{t}^{\mathrm{4}} +\mathrm{72}{t}^{\mathrm{3}} −\mathrm{74}{t}^{\mathrm{2}} −\mathrm{168}{t}−\mathrm{135}=\mathrm{0} \\ $$$$\left({t}^{\mathrm{2}} −{t}−\mathrm{9}\right)\left({t}^{\mathrm{2}} −{t}−\mathrm{3}\right)\left({t}^{\mathrm{2}} +{t}−\mathrm{5}\right)\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{but}\:\mathrm{only}\:\mathrm{2}\:\mathrm{solutions}\:\mathrm{fit} \\ $$$$\mathrm{the}\:\mathrm{original}\:\mathrm{equation} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{5}−\sqrt{\mathrm{37}}}{\mathrm{2}}\wedge{x}_{\mathrm{2}} =\frac{\mathrm{5}−\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${x}\leqslant\frac{\mathrm{5}−\sqrt{\mathrm{37}}}{\mathrm{2}}\vee{x}\geqslant\frac{\mathrm{5}−\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$
Answered by mr W last updated on 30/Apr/21
for x≤0: x^2 −4x−x≥3 x^2 −5x−3≥0 x≤((5−(√(37)))/2) for 0≤x≤4: −x^2 +4x+x≥3 x^2 −5x+3≤0 ((5−(√(13)))/2)≤x≤((5+(√(13)))/2) ⇒((5−(√(13)))/2)≤x≤4 for x≥4: x^2 −4x+x≥3 x^2 −3x−3≥0 x≥((3+(√(21)))/2) ⇒x≥4 summary: x≤((5−(√(37)))/2) or x≥((5−(√(13)))/2)
$${for}\:\boldsymbol{{x}}\leqslant\mathrm{0}: \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{x}−{x}\geqslant\mathrm{3} \\ $$$${x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{3}\geqslant\mathrm{0} \\ $$$${x}\leqslant\frac{\mathrm{5}−\sqrt{\mathrm{37}}}{\mathrm{2}} \\ $$$$ \\ $$$${for}\:\mathrm{0}\leqslant\boldsymbol{{x}}\leqslant\mathrm{4}: \\ $$$$−{x}^{\mathrm{2}} +\mathrm{4}{x}+{x}\geqslant\mathrm{3} \\ $$$${x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{3}\leqslant\mathrm{0} \\ $$$$\frac{\mathrm{5}−\sqrt{\mathrm{13}}}{\mathrm{2}}\leqslant{x}\leqslant\frac{\mathrm{5}+\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{5}−\sqrt{\mathrm{13}}}{\mathrm{2}}\leqslant{x}\leqslant\mathrm{4} \\ $$$$ \\ $$$${for}\:\boldsymbol{{x}}\geqslant\mathrm{4}: \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{x}+{x}\geqslant\mathrm{3} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{3}\geqslant\mathrm{0} \\ $$$${x}\geqslant\frac{\mathrm{3}+\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$$$\Rightarrow{x}\geqslant\mathrm{4} \\ $$$$ \\ $$$${summary}: \\ $$$${x}\leqslant\frac{\mathrm{5}−\sqrt{\mathrm{37}}}{\mathrm{2}}\:{or}\:{x}\geqslant\frac{\mathrm{5}−\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$

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