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


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

$S o l v e i n R$ $a) 3 ∣ x - \sqrt{3} ∣ - 8 \sqrt{(∣ x - \sqrt{3} ∣) + 4} = 0$ $b) \sqrt{(∣ x^{2} - x - 6} ∣) = x + 1$ $c) \sqrt{(x^{3} - 27} + 6 < x + 3$
	Answered by OlafThorendsen last updated on 19/Jul/20

	$a) u = ∣ x - \sqrt{3} ∣$ $9 u^{2} = 64 (u + 4)$ $9 u^{2} - 64 u - 256 = 0$ $Δ = 13312$ $u = \frac{64 \pm 32 \sqrt{13}}{18} = \frac{16}{9} (2 \pm \sqrt{13})$ $u > 0 \Rightarrow u = \frac{16}{9} (2 + \sqrt{13})$ $x - \sqrt{3} = \pm \frac{16}{9} (2 + \sqrt{13})$ $x = \sqrt{3} \pm \frac{16}{9} (2 + \sqrt{13})$
	Commented bymathocean1 last updated on 19/Jul/20

	$T h a n k y o u s i r!$
	Answered by bemath last updated on 20/Jul/20
	$(c) (√(x^3 −27)) < x−3 (1) x−3 > 0 →x>3 (2) x^3 −27≥0 ⇒x ≥ 3 (3)x^3 −27 < x^2 −6x+9 (x−3)(x^2 +3x+9)<(x−3)^2 (x−3){x^2 +3x+9+3−x}<0 (x−3)(x^2 +2x+12)<0 (x−3)((x+1)^2 +11)<0 x < 3 solution (1)∩(2)∩(3) x = ∅$
	$(c) \sqrt{x^{3} - 27} < x - 3$ $(1) x - 3 > 0 \to x > 3$ $(2) x^{3} - 27 ⩾ 0 \Rightarrow x ⩾ 3$ $(3) x^{3} - 27 < x^{2} - 6 x + 9$ $(x - 3) (x^{2} + 3 x + 9) < {(x - 3)}^{2}$ $(x - 3) {x^{2} + 3 x + 9 + 3 - x} < 0$ $(x - 3) (x^{2} + 2 x + 12) < 0$ $(x - 3) ({(x + 1)}^{2} + 11) < 0$ $x < 3$ $s o l u t i o n (1) \cap (2) \cap (3)$ $x = \emptyset$
	Commented bymathocean1 last updated on 20/Jul/20

	$t h a n k y o u s i r$
	Answered by Rasheed.Sindhi last updated on 20/Jul/20
	$a) 3∣x−(√3)∣−8(√((∣x−(√3)∣)+4))=0 Let (√((∣x−(√3)∣)+4))=y ∣x−(√3)∣=y^2 −4 3(y^2 −4)−8y=0 3y^2 −8y−12=0 y=((8±(√(64+144)))/3) (√((∣x−(√3)∣)+4))=((8±4(√(13)))/3) (∣x−(√3)∣)+4=(((8±4(√(13)))/3))^2 ∣x−(√3)∣=(((8±4(√(13)))/3))^2 −4 ∣x−(√3)∣=(((8±4(√(13)))/3))^2 −4 { ((x−(√3)=(((8±4(√(13)))/3))^2 −4>0)),((−x+(√3)=(((8±4(√(13)))/3))^2 −4>0)) :} { ((x=(((8±4(√(13)))/3))^2 −4+(√3))),((x=−(((8±4(√(13)))/3))^2 +4+(√3))) :}$
	$a) 3 ∣ x - \sqrt{3} ∣ - 8 \sqrt{(∣ x - \sqrt{3} ∣) + 4} = 0$ $L e t \sqrt{(∣ x - \sqrt{3} ∣) + 4} = y$ $∣ x - \sqrt{3} ∣= y^{2} - 4$ $3 (y^{2} - 4) - 8 y = 0$ $3 y^{2} - 8 y - 12 = 0$ $y = \frac{8 \pm \sqrt{64 + 144}}{3}$ $\sqrt{(∣ x - \sqrt{3} ∣) + 4} = \frac{8 \pm 4 \sqrt{13}}{3}$ $(∣ x - \sqrt{3} ∣) + 4 = {(\frac{8 \pm 4 \sqrt{13}}{3})}^{2}$ $∣ x - \sqrt{3} ∣= {(\frac{8 \pm 4 \sqrt{13}}{3})}^{2} - 4$ $∣ x - \sqrt{3} ∣= {(\frac{8 \pm 4 \sqrt{13}}{3})}^{2} - 4$ ${\begin{cases} x - \sqrt{3} = {(\frac{8 \pm 4 \sqrt{13}}{3})}^{2} - 4 > 0 \\ - x + \sqrt{3} = {(\frac{8 \pm 4 \sqrt{13}}{3})}^{2} - 4 > 0 \end{cases}$ ${\begin{cases} x = {(\frac{8 \pm 4 \sqrt{13}}{3})}^{2} - 4 + \sqrt{3} \\ x = - {(\frac{8 \pm 4 \sqrt{13}}{3})}^{2} + 4 + \sqrt{3} \end{cases}$
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