Question

解下列一元二次方程：1）(2x-3)^2=(x+1)(x-1) 2）(3-2√2)x^2+2(√2 -1)x-3=0

Answer 1

(2x-3)^2=(x+1)(x-1)
4x^2-12x+9=x^2-1
3x^2-12x+10=0
x1=(12+√144-4*3*10）/2=(12+√74）/2
x2=(12-√144-4*3*10）/2=(12-√74）/2

(3-2√2)x^2+2(√2 -1)x-3=0
(2-2√2+1)x^2+2(√2 -1)x-3=0
((√2-1)x)^2+2(√2 -1)x-3=0
((√2-1)x-1)((√2-1)x+3)=0
((√2-1)x-1)=0 ((√2-1)x+3)=0
x1=√2+1 x2=-3(√2+1)

Answer 2

1
4x^2-12x+9=x^2-1 3x^2-12x+10=0
x=(6+-√6)/3

2
[(√2-1)x]^2+2[(√2-1)x]-3=0
[(√2-1)x]=1,或[(√2-1)x]＝－3
x=√2+1 或x=-3-3√2

Answer 3

解：(1)展开4x^2-12x+9=x^2-1
合并3x^2-12x+10=0
∴a=3，b=-12，c=10
∴△=b^2-4ac=（-12）^2-4×3×10=24>0
∴x=[-b±√△]/(2a)
=[12±√24]/(2×3)
=[12±2√6]/6
=（6+√6）/3

（2）拆开(2-2√2+1)x^2+2(√2 -1)x-3=0
∴合并[(√2-1)^2]x^2+2(√2 -1)x-3=0
∴合并[(√2-1)x-1][(√2-1)x+3]=0
∴[(√2-1)x-1]=0或[(√2-1)x+3]=0
∴x1=√2+1 x2=-3√2-3