Question

用一元二次方程解：(1) 2x²-4x+1=0 (2) 2y²-7y+3=0 (3) 2x²

Answer 1

（1） 2x²-4x+1=0

x²-2x+1/2=0

x²-2x+1-1+1/2=0

(x-1)²=1/2

x-1=±√2/2

x=1±√2/2

(2)2y²-7y+3=0

(y-3)(2y-1)=0

y-3=0 或 2y-1=0,即 y=3 或 y=1/2

(3) 2x²-12x+7=0

x²-12x+36-36+7=0

(x-6)²=29

x-6=±√29

x=6±√29