Question

用换元法解一元二次方程,

Answer 1

令x^2+x=t

则原式=t(t-2)=-1

t^2-2t+1=0

(t-1)^2=0

t=1

x^2+x=1

x^2+x+1/4=1+1/4=5/4

(x+1/2)^2=5/4

x+1/2=±√5 /2

x=-1/2±√5 /2

=(-1±√5)/2