Question

resolva as equacoes abaixo em r a) ×2+5×+6=0

Answer 1

$x^{2} +5x+6=0$

Δ = $5^{2} -4.1.6=25-24=1$

$x=\frac{-5+\sqrt{1}}{2.1} =\frac{-5+1}{2} =\frac{-4}{2} =-2\\ \\ x=\frac{-5-\sqrt{1}}{2.1} =\frac{-5-1}{2} =\frac{-6}{2} =-3$

∴ S = {-2, -3}

Answer 2

x² + 5x + 6 = 0

Δ = b²−4ac

Δ = (5)² −4⋅(1)⋅(6)

Δ = 25 − 24

Δ = 1

===

$x = \dfrac{-b \pm \sqrt{\Delta}}{2*a} x = \dfrac{-5 \pm \sqrt{1}}{2*1}\\ \\ \\ x = \dfrac{-5 \pm 1}{2}\\ \\ \\ x' = \dfrac{-5 - 1}{2}\\ \\ \\ x' = \dfrac{-6}{2}\\ \\ \\ x' = -3\\ \\ \\ x'' = \dfrac{-5 + 1}{2}\\ \\ \\ x'' = \dfrac{-4}{2}\\ \\ \\ x'' = -2$

S = {-3, -2}