Przedstaw Trójmiasto kwadratowy w postaci iloczynowej jesli tomozliwe
a)y=x^2+3x-28
b)y=3x^2-7x+2
c)y=2x^2-3x+4
d)y=x^2+3x-28


Odpowiedź :

[tex]y = a(x-x_1)(x-x_2) \ - \ postac \ iloczynowa \ trojmianu \ kwadratowego[/tex]

a)

[tex]y = x^{2}+3x-28\\\\a = 1, \ b = 3, \ c = -28\\\\\Delta = b^{2}-4ac = 3^{2}-4\cdot1\cdot(-28) = 9 + 112 = 121\\\\\sqrt{\Delta} = \sqrt{121} = 11\\\\x_1=\frac{-b\sqrt{\Delta}}{2a} = \frac{-3-11}{2\cdot1} = \frac{-14}{2} = -7\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{-3+11}{2} =\frac{8}{2} = 4\\\\\boxed{y = (x+7)(x-4)} \ - \ postac \ iloczynowa[/tex]

b)

[tex]y = 3x^{2}-7x+2\\\\a = 3, \ b = -7, \ c = 2\\\\\Delta = b^{2}-4ac = (-7)^{2}-4\cdot3\cdot2 = 49-24 = 25\\\\\sqrt{\Delta} = \sqrt{25} = 5\\\\x_1 = \frac{7-5}{6} = \frac{2}{6} = \frac{1}{3}\\\\x_2 = \frac{7+5}{6} = \frac{12}{6} = 2\\\\\boxed{y = 3(x-\frac{1}{3})(x-2)} \ - \ postac \ iloczynowa[/tex]

c)

[tex]y = 2x^{2}-3x+4\\\\a = 2, \ b = -3, \ c = 4\\\\\Delta = b^{2}-4ac = (-3)^{2}-4\cdot2\cdot4 = 9-32 = -23\\\\\Delta < 0, \ brak \ pierwiastkow \ rownania, \ brak \ postaci \ iloczynowej[/tex]

d) = a)