How do you solve #2x^2-4x-3=0# by completing the square?

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Answer 1 · 2016-07-11T13:36:14

# x=1+-sqrt(5/2).#

Taking, #sqrt(5/2)~=1.581#, we get the soln. set #={2.581, -0.581}.#

Given that, #2x^2-4x-3=0.#

#:. 2x^2-4x=3.#

To complete the square on #L.H.S.#, we need
Last Term #= (Midl.Term)^2/(4xx First Term)=(-4x)^2/(4*2x^2)=(16x^2)/(8x^2)=2#

Adding #2# on both sides,

#2x^2-4x+2=3+2#

#:. 2(x^2-2x+1)=5#

#:. 2(x-1)^2=5#

#:. (x-1)^2=5/2#

#:. (x-1)=+-sqrt(5/2)#

#:. x=1+-sqrt(5/2),# is the desired soln.

Taking, #sqrt(5/2)~=1.581#, we get the soln. set #={2.581, -0.581}.#