Question

6(x 2-1) -14=5 x2+x​

Answer 1

Resposta:

$6( {x}^{2} - 1) - 14 = {5x}^{2} + x \\ 6 {x}^{2} - 6 - 14 = 5 {x}^{2} + x \\ 6 {x}^{2} - 5 {x}^{2} - x = 6 + 14 \\ {x}^{2} - x = 20 \\ {x}^{2} - x - 20 = 0$

Aplicando a fórmula de Bhaskara

${x }^{2} - x - 20 = 0 \\ \\ \frac{ - ( - 1) \frac{ + }{ - } \sqrt{ {( - 1)}^{2} - 4 \times (1) \times ( - 20) } }{2 \times (1)} = \\ \\ \frac{ + 1 \frac{ + }{ - } \sqrt{1 + 80} }{2} = \\ \\ \frac{1 \frac{ + }{ - } \sqrt{81} }{2} \\ \\ \frac{1 \frac{ + }{ - } 9}{2} = \\ \\ {x}^{1} = \frac{1 + 9}{2} = \frac{10}{2} = 5 \\ \\ {x}^{2} = \frac{1 - 9}{2} = \frac{ - 8}{2} = - 4$

Answer 2

Resposta:

V = {5; - 4}

Explicação passo-a-passo:

6(x^2 - 1) - 14 = 5x^2 + x​

6x^2 - 6 - 14 - 5x^2 - x = 0

x^2 - x - 20 = 0

a = 1; b = - 1; c = - 20

/\ = b^2 - 4ac

/\ = (-1)^2 - 4.1.(-20)

/\ = 1 + 80

/\ = 81

x = [- b +/- V/\]/2a

x = [-(-1) +/- V81]/2.1

x = [ 1 +/- 9]/2

x' = (1+9)/2 = 10/2 = 5

x" = (1-9)/2 = - 8/2 = - 4

R.:

{5; - 4}