Question

3Com a fórmula de Bhaskara, resolva as seguintes equações:
a) 6.(x2 - 1)-14 = 5x2+x
b) x2 + 4x - 12 = 0​

Answer 1

Resposta:

Explicação passo-a-passo:

a)

6 . (x² - 1 ) - 14 = 5x² + x

6x² - 6 - 14 =  5x² + x

6x² - 20 = 5x² + x

x² - x - 20 = 0

Resolvendo por Bhaskára:

Δ = b² − 4ac

Δ = (−1)² − 4⋅(1)⋅(−20)

Δ = 1 + 80

Δ = 81

$x = \dfrac{-b \pm \sqrt{\triangle}}{2a}\\ \\ \\x = \dfrac{-(-1) \pm \sqrt{81}}{2. 1} \\ \\ \\x = \dfrac{1 \pm 9 }{2} \\ \\ \\x = \dfrac{1 + 9 }{2} \\ \\ \\x = \dfrac{10}{2} \\ \\ \\x ' = 5\\ \\ \\x = \dfrac{1 - 9 }{2} \\ \\ \\x = \dfrac{-8}{2} \\ \\ \\x ' = -4$

S = ( -4 ,  5)

===

b)

x² +  4x - 12 = 0

Δ = b² − 4ac

Δ = (4)² − 4⋅(1)⋅(−12)

Δ = 16 + 48

Δ = 64

$x = \dfrac{-b \pm \sqrt{\triangle}}{2a}\\ \\ \\x = \dfrac{-(4) \pm \sqrt{64}}{2. 1} \\\\\\x = \dfrac{-4 \pm 8 }{2} \\ \\\\x = \dfrac{-4 + 8 }{2} \\ \\\\x = \dfrac{4}{2} \\ \\\\x ' = 2\\ \\\\x = \dfrac{-4 - 8 }{2} \\\\\\x = \dfrac{-12}{2} \\\\\\x ' = -5$

S = {2 , -5}