Question

9 y² - 12y+4=0
a=9
b=12
c=4

Answer 1

$9y^2 - 12y + 4 = 0$

Resolvendo por Bháskara:

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

a=9, b=−12, c=4

Δ=b²−4ac
Δ=(−12)²−4*9*4
Δ=144−144
Δ=0

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

$y = \dfrac{-(-12) \pm \sqrt{0}}{2*9} \\ \\ \\ y = \dfrac{12 \pm 0}{18} \\ \\ \\ y' = \dfrac{12 + 0}{18} \\ \\ \\ y' = \dfrac{12}{18} \\ \\ \\ y' = \dfrac{2}{3} \\ \\ \\ y'' = \dfrac{12 - 0}{18} \\ \\ \\ y'' = \dfrac{12}{18} \\ \\ \\ y'' = \dfrac{2}{3}$


S = {$\dfrac{2}{3}$}