1)
[tex]\delta =0^2- 4*1*(-9)=36\\[/tex]
[tex]x_{1} = \frac{-b-\sqrt\delta}{2a} = \frac{-6}{2} = -3\\x_{2}=\frac{-b+\sqrt{\delta}}{2a} = 3\\a=1\\[/tex]
a > 0 - ramiona skierowane w gore
x ∈ <-3;3>
2)
[tex]x^2+x-2 < 0[/tex]
[tex]\delta = 1^2-4*2*(-2) = 1+16 = 17\\x_1=\frac{-1-\sqrt{17} }{2}\\x_2=\frac{-1+\sqrt{17} }{2}[/tex]
a=1
a> 0 - ramiona skierowane w gore
x ∈ ([tex]\frac{-1-\sqrt{17}}{2}; \frac{-1+\sqrt{17}}{2}[/tex])
3)
[tex]x^2-5x+4\geq 0\\\delta = (-5)^2-4*1*4=25-16=9\\x_1=\frac{5-3}{2} = 1\\x_2=\frac{5+3}{2}=4[/tex]
a = 1 - ramiona skierowane w gore
x ∈ (-∞; 1> ∪ <4; ∞)
4)
[tex]25x^2-60x+36\leq 0\\\delta = (-60)^2-4*25*36=3600-3600 = 0\\x_0=\frac{-b}{2a} = \frac{60}{50} = \frac{6}{5} = 1\frac{1}{5} = 1,2[/tex]
a = 25 - ramiona skierowane w gore
Rozwiazaniem nierownosci jest punkt x = 1,2
5)
[tex]12x^2-11x+2>0\\\delta = (-11)^2-4*12*2=121-96=25\\x_1=\frac{11-5}{24} = \frac{6}{24} = \frac{1}{4} \\x_2 = \frac{11+5}{24} = \frac{16}{24} = \frac{2}{3} \\[/tex]
a=12 - ramiona skierowane w gore
x∈(-∞; [tex]\frac{1}{4}[/tex])∪([tex]\frac{2}{3}[/tex];∞)
