How do you solve $16x^2-12x-1=0$ by completing the square?

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Answer 1 · 2015-05-23T21:06:10

$(4x-3/2)^2 = 16x^2-12x+9/4$

So

$0 = 16x^2-12x-1 = 16x^2-12x+9/4-9/4-1$

$= (4x-3/2)^2-9/4-1$

$= (4x-3/2)^2 - 13/4$

Add $13/4$ to both ends to get:

$(4x-3/2)^2 = 13/4 = (sqrt(13)/2)^2$

So $4x-3/2 = +-sqrt(13)/2$

Add $3/2$ to both sides to get:

$4x = 3/2+-sqrt(13)/2$

Divide both sides by $4$ to get

$x=3/8+-sqrt(13)/8$

How do you solve 16x^2-12x-1=016x^2-12x-1=0 by completing the square?