用一元二次方程解下列方程(1)y=5(x-1)² (2)y=2x²-4x-1(3)y=3x²-6x+2(4)y=（x+1）（x-2）

Question

用一元二次方程解下列方程(1)y=5(x-1)² (2)y=2x²-4x-1(3)y=3x²-6x+2(4)y=（x+1）（x-2）（5）-3（x+3）（x+9）急急急急急急急急急急急急急急急急急急急急急急急急急急急！！！

Answer 1

设y为已知，可得
(1) y=5(x-1)²
(x-1)²=y/5
x-1=±√(y/5)
x=±√(y/5)+1
(2) y=2x²-4x-1
y=2(x²-2x-1/2)=2(x²-2x+1-3/2)=2(x-1)²-3
2(x-1)²=y+3
(x-1)²=(y+3)/2
x-1=±√(y+3)/2
x=±[√(y+3)/2]+1
(3) y=3x²-6x+2 y=3(x²-2x+2/3)=3(x²-2x+1-1/3)=3(x-1)²-1

3(x-1)²=y+1
(x-1)²=(y+1)/3
x-1=±√(y+1)/3
x=±[√(y+1)/3]+1
(4) y=(x+1)(x-2)=x²-x-2=x²-x+1/4-9/4=(x-1/2)²-9/4
(x-1/2)²=y+9/4
x-1/2=±√(y+9/4)
x=±[√(y+9/4)]+1/2
(5) 无等式，无解