How do you solve 4x^2-2x-5=04x2−2x−5=0 by completing the square?

1 Answer
Nghi N.
Apr 30, 2015

4x^2 - 2x - 5 = 4x^2 - 2x + 1/4 - 5 - 1/4 =4x2−2x−5=4x2−2x+14−5−14=
(2x - 1/2)^2 = 5 + 1/4 = 21/4(2x−12)2=5+14=214

a. #(2x - 1/2) = (sqr21)/2 -> 2x = 1/2 + (sqr21)/2 ->

x = 1/4 + (sqr21)/4x=14+sqr214

b. (2x - 1/2) = -(sqr 21)/2 -> 2x = 1/2 - (sqr21)/2(2x−12)=−sqr212→2x=12−sqr212 ->

x = 1/4 - (sqr21)/4x=14−sqr214

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