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Question Number 139502 by bemath last updated on 28/Apr/21

(√(x^2 +8x)) ≤ 24−(x+2)(x+6)

x2+8x⩽24−(x+2)(x+6)

Answered by TheSupreme last updated on 28/Apr/21

domain: x<−8 ∨ x>0 { ((24−(x+2)(x+6)>0→12−x^2 −8x>0)),(((12−x^2 −8x)^2 ≥(x^2 +8x))) :} { ((−4−(√(28))<x<−4+(√(28)))),(((12−x^2 −8x)^2 ≥(x^2 +8x))) :} { ((−4−(√(28))<x<−8 ∨ 0<x<−4+(√(28)))),((...)) :} x^2 +8x=u (12−u)^2 ≥u 144−25u+u^2 ≥0 u<((25−(√(625−576)))/2) ∨ x>((25+(√(625−576)))/2) u<((25−7)/2) ∨ u>((25+7)/2) u<9 ∨ x>16 a) x^2 +8x<9 ((−8−(√(64+36)))/2)<x<((−8+(√(64+36)))/2) ((−8−10)/2)<x<((−8+10)/2) −9<x<1 b) x^2 +8x>16 x^2 +8x−16>0 x<((−8−(√(64+64)))/2) ∨ x> ((−8+(√(64+64)))/2) x<−4−4(√2) ∨ x> −4+4(√2) (−∞,−4−4(√2)]∪[−9,1]∪[−4+4(√2),∞) [−4−(√(28)),−8]∪[0,−4+(√(28))] Sol: [−9,−8] ∪[0,1]

domain:x<−8∨x>0{24−(x+2)(x+6)>0→12−x2−8x>0(12−x2−8x)2⩾(x2+8x){−4−28<x<−4+28(12−x2−8x)2⩾(x2+8x){−4−28<x<−8∨0<x<−4+28...x2+8x=u(12−u)2⩾u144−25u+u2⩾0u<25−625−5762∨x>25+625−5762u<25−72∨u>25+72u<9∨x>16a)x2+8x<9−8−64+362<x<−8+64+362−8−102<x<−8+102−9<x<1b)x2+8x>16x2+8x−16>0x<−8−64+642∨x>−8+64+642x<−4−42∨x>−4+42(−∞,−4−42]∪[−9,1]∪[−4+42,∞)[−4−28,−8]∪[0,−4+28]Sol:[−9,−8]∪[0,1]

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