Question

determinar os zeros de cada uma das funções quadráticas a) y= x2,-6x +8
b)x2 -5 +6
c)-x2 -4x +12
d)-x2+6x -9​

Answer 1

Explicação passo-a-passo:

a)

$\sf x^2-6x+8=0$

$\sf \Delta=(-6)^2-4\cdot1\cdot8$

$\sf \Delta=36-32$

$\sf \Delta=4$

$\sf x=\dfrac{-(-6)\pm\sqrt{4}}{2\cdot1}=\dfrac{6\pm2}{2}$

$\sf x'=\dfrac{6+2}{2}~\Rightarrow~x'=\dfrac{8}{2}~\Rightarrow~\red{x'=4}$

$\sf x"=\dfrac{6-2}{2}~\Rightarrow~x"=\dfrac{4}{2}~\Rightarrow~\red{x"=2}$

Os zeros dessa função são 2 e 4

b)

$\sf x^2-5x+6=0$

$\sf \Delta=(-5)^2-4\cdot1\cdot6$

$\sf \Delta=25-24$

$\sf \Delta=1$

$\sf x=\dfrac{-(-5)\pm\sqrt{1}}{2\cdot1}=\dfrac{5\pm1}{2}$

$\sf x'=\dfrac{5+1}{2}~\Rightarrow~x'=\dfrac{6}{2}~\Rightarrow~\red{x'=3}$

$\sf x"=\dfrac{5-1}{2}~\Rightarrow~x"=\dfrac{4}{2}~\Rightarrow~\red{x"=2}$

Os zeros dessa função são 2 e 3

c)

$\sf -x^2-4x+12=0$

$\sf \Delta=(-4)^2-4\cdot(-1)\cdot12$

$\sf \Delta=16+48$

$\sf \Delta=64$

$\sf x=\dfrac{-(-4)\pm\sqrt{64}}{2\cdot(-1)}=\dfrac{4\pm8}{-2}$

$\sf x'=\dfrac{4+8}{-2}~\Rightarrow~x'=\dfrac{12}{-2}~\Rightarrow~\red{x'=-6}$

$\sf x"=\dfrac{4-8}{-2}~\Rightarrow~x"=\dfrac{-4}{-2}~\Rightarrow~\red{x"=2}$

Os zeros dessa função são 2 e -6

d)

$\sf -x^2+6x-9=0$

$\sf \Delta=6^2-4\cdot(-1)\cdot(-9)$

$\sf \Delta=36-36$

$\sf \Delta=0$

$\sf x=\dfrac{-6\pm\sqrt{0}}{2\cdot(-1)}=\dfrac{-6\pm0}{-2}$

$\sf x'=x"=\dfrac{-6}{-2}$

$\sf \red{x'=x"=3}$

O zero dessa função é 3