How do you solve $3x^2 – 24x + 24 = 0$ using completing the square?

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Answer 1 · 2015-07-03T13:59:07

Logical first step: divide everything by $3$

$->x^2-8x+8=0$

Completing the square would mean: taking half of the $x$ -coefficient and squaring that:
$x^2-8x+(-4)^2=x^2-8x+16=(x-4)^2$

But we still have to balance the $16$ with the $8$ we had:
$->x^2-8x+16-8=0$
$->(x-4)^2-8=0->(x-4)^2=8$

So $x-4=sqrt8=2sqrt2->x=4+2sqrt2$
Or $x-4=-sqrt8=-2sqrt2->x=4-2sqrt2$

Often written as $x_(1,2)=4+-2sqrt2$

How do you solve 3x^2 – 24x + 24 = 03x^2 – 24x + 24 = 0 using completing the square?