Question

急！！！解一元二次方程.

就一道题，拜托~~~

(a^2+a+1)(a^2-6a+1)+12a^2=0.

过程多谢！！！

Answer 1

[(a²+1)+a][(a²+1)-6a]+12a²=0
(a²+1)²-5a(a²+1)-6a²+12a²=0
(a²+1)²-5a(a²+1)+6a²=0
(a²+1-2a)(a²+1-3a)=0
(a-1)²(a²-3a+1)=0
a=1,a=(3-√5)/2,a=(3+√5)/2

Answer 2

(a² + a + 1)(a² - 6a + 1) + 12a² = 0
(a² + 1)² - 5a(a² + 1) - 6a² + 12a² = 0
(a² + 1)² - 5a(a² + 1) + 6a² = 0
(a² - 2a + 1)(a² - 3a + 1) = 0
(a - 1)²(a² - 3a + 1) = 0
a1 = a2 = 1 , a3 = (3 + √5)/2 , a4 = (3 - √5)/2