Question

calcule , se possivel , as raizes as seguintes funções do 2° grau a seguir.

a) f(x) = x² − 3x + 2

b) g(x)= 2x² + 4x − 3

c) ℎ(x) = 4x − 4x² − 1
d) m(x) = 4 + 16x²

e) p(x)= − x² − 6x

f) q(x) = 5x²

g) t(x) = 3x − x² − 3

h) u(x) = 1 − 6x + 9x²

i) v(x) =− 1 −x − x²

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Answer 1

Resposta e explicação passo a passo:

a) f(x) = x² − 3x + 2

(x - 1)(x - 2) = 0

x - 1 = 0 => x = 1

x - 2 = 0 => x = 2

Portanto, as raizes são x = 1 ou x = 2

b) g(x)= 2x² + 4x − 3

x1,2 = -b ± √b^2 - 4ac/2a

x1,2 = -4 ± √(4)^2 - 4(2)(-3)/2(2)

x1,2 = -4 ± √16 + 24/4

x1,2 = -4 ± √40/4

x1,2 = -4 ± √2x2x2x5/4

x1,2 = -4 ± 2√10/4

x1,2 = -2 ± √10/2

x1 = -2 + √10/2 e x2 = -2 - √10/2

c) ℎ(x) = 4x − 4x² − 1

x1,2 = -b ± √b^2 - 4ac/2a

x1,2 = -4 ± √(4)^2 - 4(-4)(-1)/2(-4)

x1,2 = -4 ± √16 - 16/4

x1,2 = -4 ± √0/-8

x1,2 = -4/-8

x1,2 = -4/-8

x1,2 = 1/2

d) m(x) = 4 + 16x²

16x^2 = -4

x^2 = -4/16

x^2 = -1/4

x^2 = ±√-1/4

x^2 = ±√1/4 i

x^2 = ±1/2 i

x = 1/2 ou x = -1/2

e) p(x)= − x² − 6x

x1,2 = -b ± √b^2 - 4ac/2a

x1,2 = 6 ± √(-6)^2 - 4(-1)(0)/2(-1)

x1,2 = 6 ± √36 - 0/-2

x1,2 = 6 ± 6/-2

x1,2 = 3 ± 3

x1 = 0 ou x2 = -6

f) q(x) = 5x²

5x^2 = 0

x^2 = 1/5

x = ±√1/5

g) t(x) = 3x − x² − 3

x1,2 = -b ± √b^2 - 4ac/2a

x1,2 = -3 ± √(3)^2 - 4(-1)(-3)/2(-1)

x1,2 = -3 ± √9 - 12/-2

x1,2 = -3 ± √-3/-2

x1,2 = -3 ± √3i/-2

x1 = 3 + √3i/2 ou x2 = x1 = 3 - √3i/2

h) u(x) = 1 − 6x + 9x²

x1,2 = -b ± √b^2 - 4ac/2a

x1,2 = 6 ± √(-6)^2 - 4(9)(1)/2(9)

x1,2 = 6 ± √36 - 36/18

x1,2 = 6 ± √0/18

x1,2 = 6/18

x1,2 = 1/3

i) v(x) =− 1 −x − x²

x1,2 = -b ± √b^2 - 4ac/2a

x1,2 = 1 ± √(-1)^2 - 4(-1)(-1)/2(-1)

x1,2 = 1 ± √1 - 4/-2

x1,2 = 1 ± √-3/-2

x1,2 = 1 ± √3i/-2

x1,2 = -1 + √3i/2 ou x2 = -1 - √3i/2