Question

Решите уравнение 8 класс
(a-3)x^2+ax-2a+3=0

Answer 1

$(a-3)x^2+ax-2a+3=0 \\ D=a^2-4(a-3)(3-2a)=a^2-12a+8a^2+36-24a= \\ =9a^2-36a+36=(3a-6)^2 \\ x= \frac{-a\pm(3a-6)}{2(a-3)} \\ x_1= \frac{-a-3a+6}{2(a-3)} = \frac{3-2a}{a-3} \\ x_2= \frac{-a+3a-6}{2(a-3)} = 1$