Question

3x^2 -3 (√3+3) x + 3√3 + 6 =0 como resolvo esse cálculo?​

Answer 1

Resposta:

Explicação passo-a-passo:

3x^2 - 3.(√3+3) x + 3√3 + 6 =0

3x^2 -3.(√3+3) x + 3(√3 + 2) =0

3.[x^2 - (\/3 +3)x + (\/3+2)] = 0 (:3)

x^2 - (\/3+3)x + (\/3+2) = 0

a = 1; b = - (\/3+3); c = (\/3+2)

/\= b^2 - 4ac

/\ = [-(-\/3-3)]^2 - 4.1.(\/3+2)

/\ =[ (\/3+3)^2 - 4.(\/3+2)]

/\ = (\/3)^2 + 2.3.\/3 + 9 - 4\/3 - 8

/\ = 3 + 6\/3 + 9 - 4\/3 - 8

/\ = 3+9-8 + 6\/3 - 4\/3

/\ = 4 + 2\/3

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

x = [-(-\/3-3) +/- \/(4+2\/3)]/2.1

x = [\/3+3 +/- \/(4+2\/3)]/2

R.:

x' = [\/3 + 3 + \/(4+2\/3)]/2

x" = [\/3 + 3 - \/(4-2\/3)]/2