Question

Найдите корни квадратного уравнения:
1) 3х^ - 3х + 1 = 0

 2) 5х^ + 9x + 4 = 0
 3) 36x^ - 12x + 1 = 0

Answer 1

$3x^2 - 3x + 1 = 0 \ D=9-12<0 \\ 5x^2 + 9x + 4 = 0 \ D=81-80=1 \ x_1= frac{-9-1}{10} =-1 \ x_2=frac{-9+1}{10} =-0.8 \\ 36x^2 - 12x+ 1 = 0 \ (6x-1)^2=0 \ 6x=1 \ x= frac{1}{6}$