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

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Answer 1 · 2016-07-12T22:23:28

#x = 0.781 " or " x = -1.281#

The constant (1) is already on the RHS.

We need to add in a term so that the left hand side can be factored as #(x +-?)^2#

The missing term is found by dividing the coefficient of x by 2 and then squaring the answer. The result must be added to BOTH sides.

#x +1/2x + color(red)((1/4))^2 = 1 + color(red)((1/4))^2 #

#(x+1/4)^2 = 1 1/16 = 17/16#

#x + 1/4 = +-sqrt(17/16)#

#x = +-(sqrt17)/4 -1/4 = (+-sqrt17 -1)/4#

#x = 0.781 " or " x = -1.281#