Question

二次方程：2X²-3(x-1)²=x(4x-3)，(x+4)²+8(x+4)=-12，求详细解题过程

第一个答案是：X=10分之9±根号22。第二个答案是：X= -6,-10。求详细解题过程，感谢！！

Answer 1

①2x²-3(x-1)²=x(4x-3)
2x²-3(x²-2x+1)=4x²-3x
-x²+6x-3=4x²-3x
5x²-9x+3=0
△=81-60=21
∴x=(9±√21)/10

②(x+4）²+8(x+4)=-12
(x²+8x+16)+8x+32+12=0
x²+16x+60=0
(x+8）²-4=0
x+8=±2
∴x=-6或-10

Answer 2

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