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Question Number 104178 by mathocean1 last updated on 19/Jul/20

Solve in R a) 3∣x−(√3)∣−8(√((∣x−(√3)∣)+4))=0 b)(√((∣x^2 −x−6))∣)=x+1 c)(√((x^3 −27))+6<x+3

$${Solve}\:{in}\:\mathbb{R} \\ $$ $$\left.{a}\right)\:\mathrm{3}\mid{x}−\sqrt{\mathrm{3}}\mid−\mathrm{8}\sqrt{\left(\mid{x}−\sqrt{\mathrm{3}}\mid\right)+\mathrm{4}}=\mathrm{0} \\ $$ $$\left.{b}\left.\right)\sqrt{\left(\mid{x}^{\mathrm{2}} −{x}−\mathrm{6}\right.}\mid\right)={x}+\mathrm{1} \\ $$ $$\left.{c}\right)\sqrt{\left({x}^{\mathrm{3}} −\mathrm{27}\right.}+\mathrm{6}<{x}+\mathrm{3} \\ $$

Answered by OlafThorendsen last updated on 19/Jul/20

a) u = ∣x−(√3)∣ 9u^2 = 64(u+4) 9u^2 −64u−256 = 0 Δ = 13312 u = ((64±32(√(13)))/(18)) = ((16)/9)(2±(√(13))) u>0 ⇒ u = ((16)/9)(2+(√(13))) x−(√3) = ±((16)/9)(2+(√(13))) x = (√3)±((16)/9)(2+(√(13)))

$$\left.{a}\right)\:{u}\:=\:\mid{x}−\sqrt{\mathrm{3}}\mid \\ $$ $$\mathrm{9}{u}^{\mathrm{2}} \:=\:\mathrm{64}\left({u}+\mathrm{4}\right) \\ $$ $$\mathrm{9}{u}^{\mathrm{2}} \:−\mathrm{64}{u}−\mathrm{256}\:=\:\mathrm{0} \\ $$ $$\Delta\:=\:\mathrm{13312} \\ $$ $${u}\:=\:\frac{\mathrm{64}\pm\mathrm{32}\sqrt{\mathrm{13}}}{\mathrm{18}}\:=\:\frac{\mathrm{16}}{\mathrm{9}}\left(\mathrm{2}\pm\sqrt{\mathrm{13}}\right) \\ $$ $${u}>\mathrm{0}\:\Rightarrow\:{u}\:=\:\frac{\mathrm{16}}{\mathrm{9}}\left(\mathrm{2}+\sqrt{\mathrm{13}}\right) \\ $$ $${x}−\sqrt{\mathrm{3}}\:=\:\pm\frac{\mathrm{16}}{\mathrm{9}}\left(\mathrm{2}+\sqrt{\mathrm{13}}\right) \\ $$ $${x}\:=\:\sqrt{\mathrm{3}}\pm\frac{\mathrm{16}}{\mathrm{9}}\left(\mathrm{2}+\sqrt{\mathrm{13}}\right) \\ $$

Commented bymathocean1 last updated on 19/Jul/20

Thank you sir!

$${Thank}\:{you}\:{sir}! \\ $$

Answered by bemath last updated on 20/Jul/20

(c) (√(x^3 −27)) < x−3 (1) x−3 > 0 →x>3 (2) x^3 −27≥0 ⇒x ≥ 3 (3)x^3 −27 < x^2 −6x+9 (x−3)(x^2 +3x+9)<(x−3)^2 (x−3){x^2 +3x+9+3−x}<0 (x−3)(x^2 +2x+12)<0 (x−3)((x+1)^2 +11)<0 x < 3 solution (1)∩(2)∩(3) x = ∅

$$\left({c}\right)\:\sqrt{{x}^{\mathrm{3}} −\mathrm{27}}\:<\:{x}−\mathrm{3} \\ $$ $$\left(\mathrm{1}\right)\:{x}−\mathrm{3}\:>\:\mathrm{0}\:\rightarrow{x}>\mathrm{3} \\ $$ $$\left(\mathrm{2}\right)\:{x}^{\mathrm{3}} −\mathrm{27}\geqslant\mathrm{0}\:\Rightarrow{x}\:\geqslant\:\mathrm{3}\: \\ $$ $$\left(\mathrm{3}\right){x}^{\mathrm{3}} −\mathrm{27}\:<\:{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9} \\ $$ $$\left({x}−\mathrm{3}\right)\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{9}\right)<\left({x}−\mathrm{3}\right)^{\mathrm{2}} \\ $$ $$\left({x}−\mathrm{3}\right)\left\{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{9}+\mathrm{3}−{x}\right\}<\mathrm{0} \\ $$ $$\left({x}−\mathrm{3}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{12}\right)<\mathrm{0} \\ $$ $$\left({x}−\mathrm{3}\right)\left(\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{11}\right)<\mathrm{0} \\ $$ $${x}\:<\:\mathrm{3}\: \\ $$ $${solution}\:\left(\mathrm{1}\right)\cap\left(\mathrm{2}\right)\cap\left(\mathrm{3}\right) \\ $$ $${x}\:=\:\varnothing \\ $$

Commented bymathocean1 last updated on 20/Jul/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Answered by Rasheed.Sindhi last updated on 20/Jul/20

a) 3∣x−(√3)∣−8(√((∣x−(√3)∣)+4))=0 Let (√((∣x−(√3)∣)+4))=y ∣x−(√3)∣=y^2 −4 3(y^2 −4)−8y=0 3y^2 −8y−12=0 y=((8±(√(64+144)))/3) (√((∣x−(√3)∣)+4))=((8±4(√(13)))/3) (∣x−(√3)∣)+4=(((8±4(√(13)))/3))^2 ∣x−(√3)∣=(((8±4(√(13)))/3))^2 −4 ∣x−(√3)∣=(((8±4(√(13)))/3))^2 −4 { ((x−(√3)=(((8±4(√(13)))/3))^2 −4>0)),((−x+(√3)=(((8±4(√(13)))/3))^2 −4>0)) :} { ((x=(((8±4(√(13)))/3))^2 −4+(√3))),((x=−(((8±4(√(13)))/3))^2 +4+(√3))) :}

$$\left.{a}\right)\:\mathrm{3}\mid{x}−\sqrt{\mathrm{3}}\mid−\mathrm{8}\sqrt{\left(\mid{x}−\sqrt{\mathrm{3}}\mid\right)+\mathrm{4}}=\mathrm{0} \\ $$ $${Let}\:\sqrt{\left(\mid{x}−\sqrt{\mathrm{3}}\mid\right)+\mathrm{4}}={y} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\mid{x}−\sqrt{\mathrm{3}}\mid={y}^{\mathrm{2}} −\mathrm{4} \\ $$ $$\:\:\mathrm{3}\left({y}^{\mathrm{2}} −\mathrm{4}\right)−\mathrm{8}{y}=\mathrm{0} \\ $$ $$\:\:\:\:\mathrm{3}{y}^{\mathrm{2}} −\mathrm{8}{y}−\mathrm{12}=\mathrm{0} \\ $$ $$\:\:\:\:\:\:{y}=\frac{\mathrm{8}\pm\sqrt{\mathrm{64}+\mathrm{144}}}{\mathrm{3}} \\ $$ $$\:\:\:\:\sqrt{\left(\mid{x}−\sqrt{\mathrm{3}}\mid\right)+\mathrm{4}}=\frac{\mathrm{8}\pm\mathrm{4}\sqrt{\mathrm{13}}}{\mathrm{3}} \\ $$ $$\:\:\:\:\:\:\:\left(\mid{x}−\sqrt{\mathrm{3}}\mid\right)+\mathrm{4}=\left(\frac{\mathrm{8}\pm\mathrm{4}\sqrt{\mathrm{13}}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$ $$\:\:\:\:\:\:\:\mid{x}−\sqrt{\mathrm{3}}\mid=\left(\frac{\mathrm{8}\pm\mathrm{4}\sqrt{\mathrm{13}}}{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{4} \\ $$ $$\:\:\:\:\:\:\:\mid{x}−\sqrt{\mathrm{3}}\mid=\left(\frac{\mathrm{8}\pm\mathrm{4}\sqrt{\mathrm{13}}}{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{4} \\ $$ $$\begin{cases}{{x}−\sqrt{\mathrm{3}}=\left(\frac{\mathrm{8}\pm\mathrm{4}\sqrt{\mathrm{13}}}{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{4}>\mathrm{0}}\\{−{x}+\sqrt{\mathrm{3}}=\left(\frac{\mathrm{8}\pm\mathrm{4}\sqrt{\mathrm{13}}}{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{4}>\mathrm{0}}\end{cases}\:\: \\ $$ $$\begin{cases}{{x}=\left(\frac{\mathrm{8}\pm\mathrm{4}\sqrt{\mathrm{13}}}{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{4}+\sqrt{\mathrm{3}}}\\{{x}=−\left(\frac{\mathrm{8}\pm\mathrm{4}\sqrt{\mathrm{13}}}{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{4}+\sqrt{\mathrm{3}}}\end{cases}\:\: \\ $$ $$ \\ $$

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