How do you solve #2x^2+5x-1=0# by completing the square?

1 Answer
Apr 19, 2018

Answer:

#2(x+1.25)^2-4.125=0#

Explanation:

We first take the first two terms and factor out the coefficient of #x^2#:
#(2x^2)/2+(5x)/2=2(x^2+2.5x)#

Then we divide by #x#, half the integer and square what remains:
#2(x^2/x+2.5x/x)2=2(x+2.5)#
#2(x+2,5/2)=2(x+1.25)#
#2(x+1..25)^2#

Expand out the bracket:
#2x^2+2.5x+2.5x+2(1.25^2)=2x^2+5x+3.125#

Make it equal the original equations:
#2x^2+5x+3.125+a=2x^2+5x-1#

Rearrange to find #a#:
#a=-1-3.125=-4.125#

Put in #a# to the factorised equation:
#2(x+1.25)^2-4.125=0#