Question

como resolver a equacao de 2grau (x-5) elevado a 2 =4

Answer 1

$(x - 5 )^2 = 4 \\ \\ \\ (x - 5)(x - 5) = 4 \\ \\ x^2 - 5x - 5x + 25-4 = 0 \\ \\ x^2 -10x -21 = 0$

Resolvendo por Bháskara:

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

a=1, b=−10, c=21

Δ=b²−4ac
Δ=(−10)²−4*(1)*(21)
Δ=100−84
Δ=16

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


$x = \dfrac{-(-10) \pm \sqrt{16}}{2*1} \\ \\ x = \dfrac{10 \pm 4}{2} \\ \\ x' = \dfrac{10 + 4}{2} \\ \\ x' = \dfrac{14}{2} \\ \\ x' = 7 \\ \\ x'' = \dfrac{10 - 4}{2} \\ \\ x'' = \dfrac{6}{2} \\ \\ x'' = 3$


S = {7, 3}