Question

(A) (x - 1) . (x -6)

(B) (x - 1) . (x +6)
(C) (x +1) . (x -6)
(D) (x +1.(x+6)

Answer 1

Resposta: Segue abaixo:

Explicação passo a passo:

Faremos a distributiva desses valores, ou seja multiplicaremos cada um deles pelo outro:

A)

$(x - 1) . (x - 6)\\x^{2} - 6x - 1x + 6\\x^{2} - 7x + 6 = 0$

Como resultou numa equação do segundo grau, usamos a fórmula de bháskara, que é: $\frac{-b = +-\sqrt{b^{2} - 4ac } }{2a}$

Portanto:

$\frac{-(-7)+-\sqrt{7^{2} - 4 . 1 .6} }{2 . 1} \\\frac{7 +-\sqrt{49 - 24} }{2} \\\frac{7+-\sqrt{25} }{2}\\\frac{7+-5}{2}$

Assim localizamos as duas raízes:

$x= \frac{7+5}{2} \\x = \frac{12}{2}\\x = 6$

e

$x = \frac{7-5}{2}\\x = \frac{2}{2} \\x= 1$

B)

$(x - 1) . (x + 6)\\\\x^{2} + 6x - x - 6\\\\x^{2} + 5x - 6 = 0$

bháskara novamente:

$\frac{-5)+-\sqrt{5^{2} - 4 . 1 . (-6)} }{2 . 1} \\\\\frac{-5+-\sqrt{25 + 24} }{2} \\\\\frac{-5+-\sqrt{49} }{2}\\\\\frac{-5+-7}{2}$

as raízes são:

$x =\frac{-5 - 7}{2} \\x = \frac{-12}{2} \\x = -6$

e

$x = \frac{-5 +7}{2} \\x = \frac{2}{2} \\x = 1$

C)

$(x + 1) . (x - 6)\\x^{2} - 6x + x - 6\\x^{2} - 5x - 6 = 0$

bháskara:

$\frac{-(-5)+-\sqrt{(-5)^{2} - 4 . 1 . (-6)} }{2 . 1} \\\\\frac{5+-\sqrt{25 + 24} }{2} \\\\\frac{-5+-\sqrt{49} }{2}\\\\\frac{5+-7}{2}$

as raízes:

$x = \frac{5-7}{2} \\x = \frac{-2}{2} \\x = -1$

e

$x =\frac{5+7}{2}\\\\x = \frac{12}{2}\\x = 6$

D)

$(x + 1) . (x + 6)\\x^{2} + 6x + 1x + 6\\x^{2} + 7x + 6 = 0$

bháskara:

$\frac{-7+-\sqrt{7^{2} - 4 . 1 .6} }{2 . 1} \\\frac{-7 +-\sqrt{49 - 24} }{2} \\\frac{-7+-\sqrt{25} }{2}\\\frac{-7+-5}{2}$

as raízes:

$x = \frac{-7-5}{2} \\x = \frac{-12}{2} \\x = -6$

e

$x =\frac{-7 +5}{2} \\x = \frac{-2}{2} \\x = -1$

obs: demorei pra fazer e fiz com carinho, espero ter ajudado :)