How do you solve the quadratic #y^2+2y+1=0# using any method?

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Answer 1 · 2016-08-11T16:21:06

#y=-1#

#y^2+2y+1=0#

#"Solution -1"#

#y^2+2y+1=(y+1)^2#

#(y+1)^2=0#

#y=-1#

#"Solution-2" #

#Delta=sqrt(4-4*1*1)=0#

#y_("1,2")=(-b±Delta)/(2*a)#

#"So "Delta =0 " There is one root"#

#y=(-2±0)/(2*1)#

#y=-1#