Question

alguem pod mim dizer se estar Correto ?

Answer 1

Equações de segundo grau resolve-se por Bhákara, ou fatoração:

Resolvendo por Bháskara:

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$a) \\ \\ x^2 - 4x - 5 = 0$

$x = \dfrac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$

a=1, b=−4, c=−5
Δ=b²−4ac
Δ=(−4)²−4*(1)*(−5)
Δ=16+20
Δ=36

$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a}$

$x = \dfrac{-(-4) \pm \sqrt{36}}{2*1} \\ \\ \\ x = \dfrac{4 \pm 6}{2} \\ \\ \\ x' = \dfrac{4 + 6}{2} \\ \\ \\ x' = \dfrac{10}{2} \\ \\ \\ x' = 5 \\ \\ \\ x'' = \dfrac{4 - 6}{2} \\ \\ \\ x' = \dfrac{-2}{2} \\ \\ \\ x' = -1$

S = {5, -1}

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$b) \\ \\ x^{2} + 2x - 8 = 0$

$x = \dfrac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$

a=1, b=2, c=−8

Δ=b²−4ac
Δ=(2)²−4*(1)*(−8)
Δ=4+32
Δ=36

$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a}$

$x = \dfrac{-2 \pm \sqrt{36}}{2*1} \\ \\ \\ x = \dfrac{-4 \pm 6}{2} \\ \\ \\ x' = \dfrac{-2 + 6}{2} \\ \\ \\ x' = \dfrac{4}{2} \\ \\ \\ x' = 2 \\ \\ \\ x'' = \dfrac{-2 - 6}{2} \\ \\ \\ x' = \dfrac{-8}{2} \\ \\ \\ x' = -4$

S = {2, -4 }