Question

inecuati de gradul 4.
rezolvați inecuatiile:-3x²+2x+1>sau = 0
x²‐x‐6<0
‐x²‐2x+8<sau=0
x²+5x‐6>0
‐3x²+6x>0​

Answer 1

Răspuns:

-3x^2+2x+1>sau =0

a=-3

b=2

c=1

Delta=b^2-4ac=4-4×(-3)×1=4+12=16

x1=(-b+radical din delta)/2a=(-2+4)/-6=-2/-6=3

x2=(-2-4)/-6=-6/-6=1

x^2-x-6<0

a=1

b=-1

c=-6

Delta=b^2-4ac=1-4×1×(-6)=1+24=25

x1=(-b+radical din delta)/2a=(1+5)/2=3

x2=(1-5)/2=-4/2=-2

-x^2-2x+8<sau=0

a=-1

b=-2

c=8

Delta=b^2-4ac=4-4×(-1)×8=4+32=36

x1=(-b+radical din delta) /2a=(2+6)/-2=8/-2=-4

x2=(2-6)/-2=-4/-2=2

x^2+5x-6>0

a=1

b=5

c=-6

Delta=b^2-4ac=25-4×1×(-6)=25+24=49

x1=(-b+radical din delta)/2a=(-5+7)/2=1

x2=(-5-7)/2=-12/2=-6