Question

sendo f(x)=x²-6x+8 determine o valor de x de modo que:

a) f(x) = 0

b) f(x) = 8

Answer 1

a)

$f(x) = 0\\ x^{2} - 6x + 8 = 0$

Resolvendo por Bháskara:

a = 1
b = -6
c = 8

$x = \dfrac{-b\ \pm\ \sqrt{b^{2} - 4ac}}{2a}$

$x = \dfrac{-(-6)\ \pm\ \sqrt{(-6)^{2} - 4 \cdot a \cdot 8}}{2 \cdot 1}$

$x = \dfrac{6\ \pm \sqrt{36 - 32}}{2}$

$x = \dfrac{6\ \pm\ \sqrt{4}}{2}$

$x = \dfrac{6\ \pm\ 2}{2}$


$x' = \dfrac{6 + 2}{2} = \dfrac{8}{2} = 4$

$x'' = \dfrac{6 - 2}{2} = \dfrac{4}{2} = 2$

S = {2,4}

∴x pode ser 2 ou 4.


B)

$f(x) = 8\\ x^{2} - 6x + 8 = 8$

$x^{2} - 6x + 8 - 8 = 0$

$x^{2} -6x = 0$

Usando Bháskara:

a = 1
b = -6
c = 0

$x = \dfrac{-b\ \pm\ \sqrt{b^{2} - 4ac}}{2a}$

$x = \dfrac{-(-6)\ \pm\ \sqrt{(-6)^{2} - 4 \cdot 1 \cdot 0}}{2 \cdot 1}$

$x = \dfrac{6\ \pm\ \sqrt{36 - 0}}{2}$

$x = \dfrac{6\ \pm\ 6}{2}$


$x' = \dfrac{6+6}{2} = \dfrac{12}{2} = 6$

$x'' = \dfrac{6-6}{2} = \dfrac{0}{2} = 0$

S = {0,6}

∴x pode ser 6 ou 0.