Odpowiedź:
[tex]a)\\\\-x^2>3-4x\\\\-x^2+4x-3>0\\\\\Delta=b^2-4ac\\\\\Delta=4^2-4\cdot(-1)\cdot(-3)=16-12=4\\\\\sqrt{\Delta}=\sqrt{4}=2\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-4-2}{2\cdot(-1)}=\frac{-6}{-2}=3\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-4+2}{2\cdot(-1)}=\frac{-2}{-2}=1[/tex]
[tex]b)\\\\x^2\geq -5x-4\\\\x^2+5x+4\geq 0\\\\\Delta=b^2-4ac\\\\\Delta=5^2-4\cdot1\cdot4=25-16=9\\\\\sqrt{\Delta}=\sqrt{9}=3\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-5-3}{2\cdot1}=\frac{-8}{2}=-4\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-5+3}{2\cdot1}=\frac{-2}{2}=-1[/tex]
[tex]c)\\\\-x^2<-6x+8\\\\-x^2+6x-8<0\\\\\Delta=b^2-4ac\\\\\Delta=6^2-4\cdot(-1)\cdot(-8)=36-32=4\\\\\sqrt{\Delta}=\sqrt{4}=2\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-6-2}{2\cdot(-1)}=\frac{-8}{-2}=4\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-6+2}{2\cdot(-1)}=\frac{-4}{-2}=2[/tex]
[tex]d)\\\\x^2\geq 2-x\\\\x^2+x-2\geq 0\\\\\Delta=b^2-4ac\\\\\Delta=1^2-4\cdot1\cdot(-2)=1+8=9\\\\\sqrt{\Delta}=\sqrt{9}=3\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-1-3}{2\cdot1}=\frac{-4}{2}=-2\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-1+3}{2\cdot1}=\frac{2}{2}=1[/tex]
