Question

determine se houver as raizes das funçoes de r em r dadas por: x2-x-12=0
6x2+x-1=0
7x2+2x+1=0
x2-12+36=0

Answer 1

x2-x-12=0

Δ = b^2 - 4ac

Δ = (-1)^2 - 4 x 1 x (-12) = 1+48=49

x = [-b+/- raiz(Δ)]/2a

x =[ 1 +/- 7]/2x1

x' = 4    x" = -3

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6x2+x-1=0

Δ = (1)^2 - 4 x 6 x (-1) = 1+24 =25

x = [-1 +/- raiz (25)]/2x6

x' = 4/12 = 1/3  x" = -6/12 = -1/2

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7x2+2x+1=0

Δ = 2^2 - 4 x 7 x 1 = 4 -28 = -24

nao possui raizes reais

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x2-12+36=0

Δ = (-12)^2 - 4 x 1 x 36 = 144-144=0



x = -b/2a = -(-12)/2x1 = 6