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

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)))

Commented bymathocean1 last updated on 19/Jul/20

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 = ∅

Commented bymathocean1 last updated on 20/Jul/20

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))) :}