How do you solve #4x^2+5=10x# by completing the square?

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Answer 1 · 2017-06-04T16:08:50

#x = 1.809 or x = 0.691#

Re-arrange the equation to be equal to #0#.

#4x^2 -10x+5=0" "larr# make #1x^2rarr div 4#

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

#x^2 -5/2x color(white)(........)=color(red)(-5/4)#

#x^2 -5/2x color(blue)(+25/16)= -5/4color(blue)(+25/16)" "larr#Add# color(blue)((b/2)^2)# to both sides#

#(x-5/4)^2 color(white)(....)= 5/16#

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

#x = +sqrt(5/16)+5/4 = 1.809#

#x = -sqrt(5/16) +5/4 = 0.691#