Question

2) Vamos determine, no conjunto R, o conjunto solução das seguintes equações do 2º grau do 1º caso:

a) x² - 6x = 0
b) 2x² - 4x = 0
c) 8x² - 6x = 0
d) x² + 15x = 0

Answer 1

Explicação passo-a-passo:

a )

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

=> Usando a fórmula de bhaskara

$\sf ∆ = {b}^{2} - 4ac$

$\sf ∆ = {-6}^{2} - 4(1)(0)$

$\sf ∆ = 36 - 0$

$\sf ∆ = 36$

=>Valor de X

$\sf x = \frac{ - ( - 6) ± \sqrt{36} }{2}$

$\sf x = \frac{6 ± 6}{2}$

$\sf x_{1} = \frac{6 + 6}{2} = > \frac{12}{2} = > \textcolor{red}{6}$

$\sf x_{2} = \frac{6 - 6}{2} = > \frac{0}{2} = > \textcolor{red}{0}$

$S = \{ 0 \: ;\: 6 \}$

b )

$\sf 2 {x}^{2} - 4x = 0$

=>Usando a fórmula de bhaskara

$\sf ∆ = {b}^{2} - 4ac$

$\sf ∆ = {-4}^{2} - 4(2)(0)$

$\sf ∆ = 16 - 0$

$\sf ∆ = 16$

=>Valor de X

$\sf x = \frac{ - ( - 4) ± \sqrt{16} }{4}$

$\sf x = \frac{4 ± 4}{4}$

$\sf x_{1} = \frac{4 + 4}{4} = > \frac{8}{4} = > \textcolor{red}{2}$

$\sf x_{2} = \frac{4 - 4}{4} = > \frac{0}{4} = > \textcolor{red}{0}$

$S = \{ 0 \: ; \: 2 \}$

c )

$\sf 8x {}^{2} - 6x = 0$

=>Usando a fórmula de bhaskara

$\sf ∆ = {b}^{2} - 4ac$

$\sf ∆ = {-6}^{2} - 4(2)(0)$

$\sf ∆ = 36 - 0$

$\sf ∆ = 36$

=> Valor de X

$\sf x = \frac{ - ( - 6) ± \sqrt{36} }{16}$

$\sf x = \frac{6 ± 6}{16}$

$\sf x_{1} = \frac{6 + 6}{16} = > \frac{12}{16} = > \textcolor{red}{\frac{3}{4}}$

$\sf x_{2} = \frac{6 - 6}{16} = > \frac{0}{16} = > \textcolor{red}{0}$

$S = \{ 0 \: ; \: \frac{3}{4}\: \}$

d )

$\sf {x}^{2} - 15x = 0$

=>Usando a fórmula de bhaskara

$\sf ∆ = {b}^{2} - 4ac$

$\sf ∆ = {-15}^{2} - 4(1)(0)$

$\sf ∆ = 225 - 0$

$\sf ∆ = 225$

=>Valor de X

$\sf x = \frac{ - ( - 15) ± \sqrt{225} }{2}$

$\sf x = \frac{15 ± 15}{2}$

$\sf x_{1} = \frac{15 + 15}{2} = > \frac{30}{2} = > \textcolor{red}{15}$

$\sf x_{2} = \frac{15 - 15}{2} = > \frac{0}{2} = \textcolor{red}{0}$

$S = \{ 0 \: ; \: 15 \}$

Espero ter ajudado !!!