a)
[tex]x^{2} -4x + 4 = 0\\\\a = 1, \ b = -4, \ c = 4\\\\\Delta = b^{2}-4ac=(-4)^{2}-4\cdot1\cdot4 = 16-16 = 0\\\\x_{o} = \frac{-b}{2a} = \frac{-(-4)}{2}\\\\\underline{x_{o} = -2}[/tex]
[tex]Lub:[/tex]
[tex]x^{2}-4x +4 = 0\\\\(x-2)^{2} = 0\\\\x-2 = 0\\\\\underline{x_{o} = 2}[/tex]
b)
[tex]x^{2}-4x+7 = 0\\\\a = 1, \ b = -4, \ c = 7\\\\\Delta = b^{2}-4ac = (-4)^{2}-4\cdot1\cdot7 = 16-28 = -12\\\\\Delta < 0, \ brak \ rozwiazan[/tex]
c)
[tex]-4x^{2}+6x-1 = 0 \ \ /\cdot(-1)\\\\4x^{2}-6x+1 = 0\\\\a = 4, \ b = -6, \ c = 1\\\\\Delta = b^{2}-4ac = (-6)^{2}-4\cdot4\cdot1 = 36 -12 = 20\\\\\sqrt{\Delta} = \sqrt{20} = \sqrt{4\cdot5} = 2\sqrt{5}\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{6-2\sqrt{5}}{8} = \frac{3-\sqrt{5}}{4}}\\\\x_2 = \frac{-b+\sqer{\Delta}}{2a} = \frac{6+2\sqrt{5}}{8} =\frac{3+\sqrt{5}}{4}}[/tex]
d)
[tex]-4x^{2}+6x-3 = 0 \ \ /\cdot(-1)\\\\4x^{2}-6x+3 = 0\\\\a = 4, \ b = -6, \ c = 3\\\\\Delta = b^{2}-4ac = (-6)^{2}-4\cdot4\cdot3 = 36-48=-12\\\\\Delta < 0, \ brak \ rozwiazan[/tex]
Wyjaśnienie:
Gdy Δ > 0, równanie kwadratowe ma dwa rozwiązania.
Gdy Δ = 0, równanie kwadratowe ma jedno rozwiązanie.
Gdy Δ < 0, równanie kwadratowe nie ma rozwiązań.
