zad. 5
a)
[tex](2x-5)^2-(x+3)(2x-5)\geq 0\\4x^2+20x+25-(2x^2-5x+6x-15)\geq 0\\4x^2+20x+25-2x^2+5x-6x+15\geq 0\\2x^2+19x+40\geq 0\\\delta = 19^2-4*2*40\\\delta=361-320\\\delta=41\\\sqrt{\delta} = \sqrt{41}\\x_1 = \frac{-19-\sqrt{41}}{4}\\x_2 = \frac{-19+\sqrt{41}}{4}\\a > 0\\[/tex]
x ∈ (-∞; [tex]\frac{-19-\sqrt{41}}{4}[/tex]> ∪ <[tex]\frac{-19+\sqrt{41}}{4}[/tex]; ∞)
b)
[tex]x^2-4x\geq -4\\x^2-4x+4\geq 0\\\delta=16-4*1*4\\\delta=16-16\\\delta=0\\x_0=\frac{-b}{2a}\\x_0=\frac42\\x_0=2\\a > 0[/tex]
x ∈ (-∞; ∞)
c)
[tex](x+3)^2-9=0\\x^2+6x+9-9=0\\x^2+6x=0\\\delta=36-4*1*0\\\delta=36-0\\\delta=36\\\sqrt{\delta}=6\\x_1 = \frac{-6-6}{2} = \frac{-12}{2}=-6\\x_2 = \frac{-6+6}{2} = 0[/tex]
