Question

一元二次方程：（1）(3x-1)²-4(2x+3)²=0 （2）2（t-1）²+t=1

Answer 1

（1）(3x-1)²-4(2x+3)²=0
（3x-1）²-(4x+6)²=0
(3x-1+4x+6)(3x-1-4x-6)=0
(7x+5)(-x-7)=0
x1=-5/7 x2=-7

（2）2（t-1）²+t=1
2(t-1)²+t-1=0
(2t-2+1)(t-1)=0
(2t-1)(t-1)=0
t1=1/2 t2=1

Answer 2

-7x²-54x-35=0 及(-7x-5)(x+7)=0 解得x= -5/7 或 -7

2t²-3t+1=0 及 (2t-1)(t-1) 解得t=1/2 或 1

Answer 3

1.(3x-1)^2=(4x+6)^2
得：3x-1=4x+6 或者 3x-1=-(4x+6)
x1=-5/7
x2=-1

2. 2(t-1)^2+(t-1)=0
(t-1)(2t-2+1)=0
t1=1
t2=0.5

Answer 4

(3x-1)²-4(2x+3)²=0
9x^2-6x+1-(16x^2+48x+36)=0
7x^2+42x+35=0
(x+5)(7x+7)=0
x=-5 or -1

2（t-1）²+t=1
2t^2-3t+1=0
(2t-1)(t-1)=0
t=1/2 or 1