How do you solve the equation $3x^2-24x+27=0$ by completing the square?

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Answer 1 · 2017-07-04T11:55:04

You may start by dividing everything by $3$

$->x^2-8x+9=0$

You take half of the $x$ coefficient and square that. The square would then be:
$x^2-8x+(-4)^2=x^2-8x+16=(x-4)^2$

To get there you would have added $7$ to the original $9$ . Also add $7$ to the other side of the $=$ sign:

$->(x-4)^2=7->x-4=+-sqrt7->x=4+-sqrt7$

How do you solve the equation 3x^2-24x+27=03x^2-24x+27=0 by completing the square?