Rozwiąż równania:
a) 4x^2 - 121 = 0
b) x^2 - 4x + 4 = 0
c) x^2 + 8x + 16 = 0
d) x^2 - x + 1/4 = 0
e) x^2 - x + 1/4 =
f) x^2 + 4 = 0



Odpowiedź :

Magda

Odpowiedź:

[tex]a)\\\\4x^2-121=0\\\\4x^2=121\ \ |:4\\\\x^2=\frac{121}{4}\\\\x=\sqrt{\frac{121}{4}}\ \ \ \ lub\ \ \ \ x=-\sqrt{\frac{121}{4}}\\\\x=\frac{11}{2}\ \ \ \ \ \ lub\ \ \ \ x=-\frac{11}{2}\\\\\\b)\\\\x^2-4x+4=0\\\\(x-2)^2=0\\\\x-2=0\\\\x=2\\\\lub\\\\x^2-4x+4=0\\\\\Delta=b^2-4ac=(-4)^2-4\cdot1\cdot4=16-16=0\\\\\sqrt{\Delta}=0\\\\x_{0}=\frac{-b}{2a}=\frac{-(-4)}{2\cdot1}=\frac{4}{2}=2\\\\\\c)\\\\x^2+8x+16=0\\\\(x+4)^2=0\\\\x+4=0\\\\x=-4[/tex]

[tex]lub\\\\x^2+8x+16=0\\\\\Delta=b^2-4ac=8^2-4\cdot1\cdot16=64-64=0\\\\\sqrt{\Delta}=0\\\\x_{0}=\frac{-b}{2a}=\frac{-8}{2\cdot1}=\frac{-8}{2}=-4[/tex]

[tex]d)\\\\x^2-x+\frac{1}{4}=0\ \ |\cdot4\\\\4x^2-4x+1=0\\\\(2x-1)^2=0\\\\2x-1=0\\\\2x=1\ \ |:2\\\\x=\frac{1}{2}\\\\lub\\\\x^2-x+\frac{1}{4}=0\ \ |\cdot4\\\\4x^2-4x+1=0\\\\\Delta=b^2-4ac=(-4)^2-4\cdot4\cdot1=16-16=0\\\\\sqrt{\Delta}=0\\\\x_{0}=\frac{-b}{2a}=\frac{-(-4)}{2\cdot4}=\frac{4}{8}=\frac{1}{2}[/tex]

e) przykład podobny do d)

[tex]f)\\\\x^2+4=0\\\\x^2=-4\\\\R\'ownanie\ \ nie\ \ ma\ \ rozwiazania[/tex]