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

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

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Answer 1 · 2017-07-04T11:55:04

You may start by dividing everything by $3$ $3$

Explanation:

$\to x^{2} - 8 x + 9 = 0$ $->x^2-8x+9=0$

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

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

$\to {(x - 4)}^{2} = 7 \to x - 4 = \pm \sqrt{7} \to x = 4 \pm \sqrt{7}$ $->(x-4)^2=7->x-4=+-sqrt7->x=4+-sqrt7$

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

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How do you solve the equation 3x2−24x+27=03x^2-24x+27=0 by completing the square?

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