Question

2(t-1)²+8t=0用因式分解法怎么计算？谢谢

Answer 1

原方程两边都除以2，得
(t-1)²+4t=0
等式左边展开、合并同类项，得(t+1)²=0
所以t+1=0
所以t=-1

Answer 2

2(t-1)²+8t=0
2（t²-2t+1）+8t=0
2t²-4t+2+8t=0
2t²+4t+2=0
2（t²+2t+1）=0
2（t+1）²=0