Question #35aac

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Answer 1 · 2018-01-05T13:54:07

Assuming you meant to find solutions for $1/2x^2-4x+4=0$ ,

$x= +-2sqrt(2) +4$

Firstly, take out the $1/2$ term and complete the square as usual

$1/2x^2-4x+4 = 0$
$1/2[x^2-8x+8] = 0$
$1/2[(x-4)^2 -8] = 0$
$1/2(x-4)^2 -4 = 0$

Now solve for $x$ ,

$1/2(x-4)^2= 4$
$(x-4)^2 = 8$
$x-4= +-sqrt(8)$
$x= +-sqrt(8)+4$

And simplify the root term

$x= +-(sqrt(4) * sqrt(2)) +4$
$x= +-2sqrt(2) +4$