Question

Determine os zeros ou raízes de cada uma das funções quadráticas.
e) Y= -2x²+6x-4
f) Y= 3x²-4x+1
g) Y= -x²+6x-0
h) Y= -4x²-8x-4
me ajudemmmmm pfv

Answer 1

Resposta:

$e:\ x=1\ ou\ x=2\\\\f:\ x=1\ ou\ x=\frac{1}{3}\\\\g:\ x=0\ ou\ x=6\\\\h:\ x=-1$

Explicação passo a passo:

$e)\ y=-2x^2+6x-4\\a=-2,\ b=6\ e\ c=-4\\\\x_{1,2}=\frac{-b\±\sqrt{b^2-4ac}}{2a}\\\\x_{1,2}=\frac{-6\±\sqrt{6^2-4(-2)(-4)}}{2(-2)}\\\\x_{1,2}=\frac{-6\±\sqrt{36-32}}{-4}\\\\x_{1,2}=\frac{-6\±\sqrt{4}}{-4}\\\\x_{1,2}=\frac{-6\±2}{-4}$

$x_{1}=\frac{-6+2}{-4}=\frac{-4}{-4}=1\\\\x_{2}=\frac{-6-2}{-4}=\frac{-8}{-4}=2$

$f)\ y=3x^2-4x+1\\a=3,\ b=-4\ e\ c=1\\\\x_{1,2}=\frac{-b\±\sqrt{b^2-4ac}}{2a}\\\\x_{1,2}=\frac{-(-4)\±\sqrt{(-4)^2-4(3)(1)}}{2(3)}\\\\x_{1,2}=\frac{4\±\sqrt{16-12}}{6}\\\\x_{1,2}=\frac{4\±\sqrt{4}}{6}\\\\x_{1,2}=\frac{4\±2}{6}\\\\x_{1}=\frac{4+2}{6}=\frac{6}{6}=1\\\\x_{2}=\frac{4-2}{6}=\frac{2}{6}=\frac{1}{3}$

$g)\ y=-x^2+6x\\a=-1,\ b=6\ e\ c=0\\\\x_{1,2}=\frac{-b\±\sqrt{b^2-4ac}}{2a}\\\\x_{1,2}=\frac{-6\±\sqrt{6^2-4(-1)(0)}}{2(-1)}\\\\x_{1,2}=\frac{-6\±\sqrt{36}}{-2}\\\\x_{1,2}=\frac{-6\±6}{-2}\\\\x_{1}=\frac{-6+6}{-2}=\frac{0}{-2}=0\\\\x_{1,2}=\frac{-6-6}{-2}=\frac{-12}{-2}=6$

$h)\ -4x^2-8x-4\\a=-4,\ b=-8\ e\ c=-4\\\\x_{1,2}=\frac{-b\±\sqrt{b^2-4ac}}{2a}\\\\x_{1,2}=\frac{-(-8)\±\sqrt{(-8)^2-4(-4)(-4)}}{2(-4)}\\\\x_{1,2}=\frac{8\±\sqrt{64-64}}{-8}\\\\x_{1,2}=\frac{8\±\sqrt{0}}{-8}\\\\x_{1,2}=\frac{8}{-8}=x_{1,2}=-1$