b)
\frac{1}{4}x^2-\frac{x}{3}+\frac{1}{9}=0/*36
9x^2-12x+4=0
a=9, b=-12, c=4
\Delta=144-4*9*4=144-144=0
x_0=\frac{-b}{2a}=\frac{12}{2*9}=\frac{12}{18}=\frac{2}{3} pierwiastek dwukrotny
c)
(\sqrt2-1)x^2+2x+2=0
a=\sqrt2-1 , b=2 , c=2
\Delta=b^2-4ac=4-4*(\sqrt2-1)*2=4-8(\sqrt2-1)=4-8\sqrt2+8=12-8\sqrt2
\sqrt\Delta=\sqrt{12-8\sqrt2}=\sqrt{4(3-2\sqrt2)}=2\sqrt{3-2\sqrt2}=2\sqrt{2-2\sqrt2+1}=2\sqrt{(\sqrt2-1)^2}=2(\sqrt2-1)=2\sqrt2-2
x_1=\frac{-b-\sqrt\Delta}{2a}=\frac{-2-(2\sqrt2-2)}{2(\sqrt2-1)}=\frac{-2-2\sqrt2+2}{2(\sqrt2-1)}=\frac{-2\sqrt2}{2(\sqrt2-1)}=\frac{-\sqrt2}{\sqrt2-1}=\frac{-\sqrt2(\sqrt2+1)}{(\sqrt2-1)(\sqrt2+1)}=\frac{-2-\sqrt2}{2-1}=-2-\sqrt2
x_2=\frac{-b+\sqrt\Delta}{2a}=\frac{-2+2\sqrt2-2}{2(\sqrt2-1)}=\frac{2\sqrt2-4}{2(\sqrt2-1)}=\frac{2(\sqrt2-2)}{2(\sqrt2-1)}=\frac{\sqrt2-2}{\sqrt2-1}=\frac{(\sqrt2-2)(\sqrt2+1))}{(\sqrt2-1)(\sqrt2+1)}=\frac{2+\sqrt2-2\sqrt2-2}{2-1}=-\sqrt2
d)
x^4-16x^2=0
x^2(x^2-16)=0
x^2(x-4)(x+4)=0
x^2=0\vee x-4=0\vee x+4=0
x=0\vee x=4\vee x=-4