Odpowiedź :
Postać iloczynowa trójmianu kwadratowego to postać:
[tex]y=a(x-x_1)(x-x_2)[/tex], gdzie [tex]x_1[/tex] i [tex]x_2[/tex] są miejscami zerowymi funkcji, a [tex]a[/tex] jest współczynnikiem przy najwyższej potędze [tex]x[/tex] z postaci ogólnej.
a)
[tex]2x^2-5x+2=0\\\Delta=b^2-4ac=\left(-5\right)^2-4*2*2=25-16=9\\x_1=\cfrac{-b-\sqrt\Delta}{2a}=\cfrac{5-\sqrt{9}}{2*2}=\cfrac{5-3}{4}=0.5\\x_2=\cfrac{-b+\sqrt\Delta}{2a}=\cfrac{5+\sqrt{9}}{2*2}=\cfrac{5+3}{4}=2\\y=2(x-0.5)(x-2)[/tex]
b)
[tex]-x^2+2\sqrt{3}x+1=0\\\Delta=b^2-4ac=\left(2\sqrt{3}\right)^2-4*\left(-1\right)*1=12+4=16\\x_1=\cfrac{-b-\sqrt\Delta}{2a}=\cfrac{-2\sqrt{3}-\sqrt{16}}{2*\left(-1\right)}=\cfrac{-2\sqrt{3}-4}{\left(-2\right)}=\sqrt{3}+2\\x_2=\cfrac{-b+\sqrt\Delta}{2a}=\cfrac{-2\sqrt{3}+\sqrt{16}}{2*\left(-1\right)}=\cfrac{-2\sqrt{3}+4}{\left(-2\right)}=\sqrt{3}-2\\y=-(x-\sqrt{3}-2)(x-\sqrt{3}+2)[/tex]
c)
[tex]-3x^2-x+2=0\\\Delta=b^2-4ac=\left(-1\right)^2-4*\left(-3\right)*2=1+24=25\\x_1=\cfrac{-b-\sqrt\Delta}{2a}=\cfrac{1-\sqrt{25}}{2*\left(-3\right)}=\cfrac{1-5}{\left(-6\right)}=\cfrac{2}{3}\\x_2=\cfrac{-b+\sqrt\Delta}{2a}=\cfrac{1+\sqrt{25}}{2*\left(-3\right)}=\cfrac{1+5}{\left(-6\right)}=-1\\y=-3\left(x-\cfrac{2}{3}\right)(x+1)[/tex]
d)
[tex]\cfrac{1}{2}x^2-\sqrt{5}x+2=0\\\Delta=b^2-4ac=\left(-\sqrt{5}\right)^2-4*\cfrac{1}{2}*2=5-4=1\\x_1=\cfrac{-b-\sqrt\Delta}{2a}=\cfrac{\sqrt{5}-\sqrt{1}}{2*\left(\frac{1}{2}\right)}=\cfrac{\sqrt{5}-1}{1}=\sqrt{5}-1\\x_2=\cfrac{-b+\sqrt\Delta}{2a}=\cfrac{\sqrt{5}+\sqrt{1}}{2*\left(\frac{1}{2}\right)}=\cfrac{\sqrt{5}+1}{1}=\sqrt{5}+1\\y=\cfrac{1}{2}(x-\sqrt{5}+1)(x-\sqrt{5}-1)[/tex]