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

1 Answer
Harold Walden
Jan 12, 2018

This question doesn't have any real solutions. However it does have two complex solutions;
#color(red)(x=(1+-isqrt(39))/4)#

Explanation:

#2x^2-x=-5#
#2x^2-x+5=0#
#2(x^2-x/2+5/2)=0#
#x^2-x/2+5/2=0#
#(x-1/4)^2-1/16+5/2=0#
#(x-1/4)^2+39/16=0#
#(x-1/4)^2=-39/16#
#x-1/4=+-sqrt(-39/16)#
#x-1/4=+-sqrt(-39)/4=+-(sqrt(39)xxsqrt(-1))/4#
#x=1/4+-(sqrt(39)xxsqrt(-1))/4=1/4+-(sqrt(39)xxi)/4#
#color(red)(x=(1+-isqrt(39))/4)#

I hope it helps :)

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