Question

rezolvati in r inecuatia: a) (x-4)^2+12≥(3x-2)^2,,,b) 3x(x+ radical3)≤(x+radical3)^2
AJUTOR VA ROG FOARTE MULT

Answer 1

a) Desfaci parantezele

x^2-8x+16+12 >= 9x^2-12x+4

trecem tot in stanga

-8x^2+4x+24 >=0

delta : 16+4*8*24=784
radical din delta = 28

x1 = (-4+28)/-16=24/-16=-1.5
x2= 32/16=2

intre radacini avem semn contrar lui a deci solutia este [-1.5,2]

b)

3x^2+3radical(3)x <=x^2+2radical(3)x+3
2x^2+radical(3)x-3 <= 0

delta : 3+24=27
radical din delta =3radical(3)

x1 = (-radical(3)+3radical(3))/4=radical(3)/2
x2=(-radical(3)-3radical(3))/4=-radical(3)

deci solutia este [-radical(3),radical(3)/2]