Question

x²+10x+24=0 em forma de baskara e delta x²-2x-3=0 em forma de baskara e delta

Answer 1

x² + 10x + 24 = 0

Δ = b² − 4ac

Δ = (10)² −4⋅(1)⋅(24)

Δ = 100 − 96

Δ = 4

$x = \dfrac{-b \pm \sqrt{\Delta}}{2*a}\\ \\ \\ x = \dfrac{-10 \pm \sqrt{4}}{2*1}\\ \\ \\ x = \dfrac{-10 \pm 2}{2}\\ \\ \\ x' = \dfrac{-10 - 2}{2}\\ \\ \\ x' = \dfrac{-12}{2}\\ \\ \\ x' = -6\\ \\ \\ x'' = \dfrac{-10 + 2}{2}\\ \\ \\ x'' = \dfrac{-8}{2}\\ \\ \\ x'' = -4$

S = {-6, -4}

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x² -2x - 3 = 0

Δ = b²−4ac

Δ = (−2)² −4⋅(1)⋅(−3)

Δ = 4 + 12

Δ = 16

$x = \dfrac{-b \pm \sqrt{\Delta}}{2*a}\\ \\ \\ x = \dfrac{-(-2) \pm \sqrt{16}}{2*1}\\ \\ \\ x = \dfrac{2 \pm 4}{2}\\ \\ \\ x' = \dfrac{2 - 4}{2}\\ \\ \\ x' = \dfrac{-2}{2}\\ \\ \\ x' = -1\\ \\ \\ x'' = \dfrac{2 + 4}{2}\\ \\ \\ x'' = \dfrac{6}{2}\\ \\ \\ x'' = 3$

S = {-1, 3}

Answer 2

x² + 10x + 24 = 0

a = 1

b = 10

c = 24

Δ = b² - 4ac

Δ = 10² - 4 . 1 . 24

Δ = 100 - 96

Δ = 4

x = (- b +- √Δ)/2a

x = (- 10 +- √4)/2 . 1

x = (- 10 +- 2)/2

x' = (- 10 + 2)/2

x' = - 8/2

x' = - 4

x'' = (- 10 - 2)/2

x'' = - 12/2

x'' = - 6

.............................................................................

x² - 2x - 3 = 0

a = 1

b = - 2

c = - 3

Δ = b² - 4ac

Δ = (- 2)² - 4 . 1 . (-3)

Δ = 4 + 12

Δ = 16

x = (- b +- √Δ)/2a

x = (+ 2 +- √16)/2 . 1

x = (+ 2 +- 4)/2

x' = (+ 2 + 4)/2

x' = 6/2

x' = 3

x'' = (+ 2 - 4)/2

x'' = - 2/2

x'' = - 1