How do I solve $4x^2 - 6x + 1 = 0$ by completing the square?

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Answer 1 · 2014-09-13T22:54:49

$4x^2-6x=-1$

Factor out a $4$

$4(x^2-6/4x)=-1$

Simplify fraction

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

Take half of x term and square the result

$((-3/2)/2)^2=((-3/2)*(1/2))^2=(-3/4)^2=9/16$

Add $4*9/16$ or $9/4$ to the right side because we factored out a $4$ .
Add $9/16$ to the left side.

$4(x^2-3/2x+9/16)=-1+9/4$

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

$4(x-3/4)^2=-4/4+9/4$ , common denominators $-1=-4/4$

$4(x-3/4)^2=5/4$

$(x-3/4)^2=(5/4)/4$

$(x-3/4)^2=5/16$

$sqrt((x-3/4)^2)=+-sqrt(5/16)$

$x-3/4=+-sqrt(5/16)$

$x-3/4=+-sqrt(5)/4$

$x=+-sqrt(5)/4+3/4$

$x=(+-sqrt(5)+3)/4$

How do I solve 4x^2 - 6x + 1 = 04x^2 - 6x + 1 = 0 by completing the square?