How do you use completing the square method to solve #4x^2+5x=-1#?

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Answer 1 · 2018-03-24T06:05:02

#x =-1/4#
#x=-1#

#4x^2+5x=-1#

#4x^2+5x+1=0#

#(2x)^2+5x+1=0#

#(2x)^2+2×(2x)×(5/4)+(5/4)^2-(5/4)^2+1=0#

#(2x+5/4)^2 -(5/4)^2+1=0#

#(2x+5/4)^2 =25/16-1#

#(2x+5/4)^2 =9/16#

#2x+5/4=+-sqrt(9/16)#

#2x+5/4=3/4 or 2x+5/4=-3/4#

#2x=3/4-5/4 or 2x= -3/4-5/4#

#2x=-2/4 or 2x= -8/4#

#x=-1/4 or x=-1#