How do I solve $x^{2} + 12 x = - 27$ $x^2 + 12x = -27$ by completing the square?

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Answer 1 · 2014-08-29T12:39:14

$x^{2} + 12 x = - 27$ $x^2+12x=-27$

Take the co-efficient of the middle $x$ $x$ term and half it

$\frac{12}{2} = 6$ $12/2 =6$

Square that number and add and subtract it from the equation
This gives the perfect square component.

$6^{2} = 36$ $6^2=36$

$x^{2} + 12 x + (36) - (36) = - 27$ $x^2+12x+(36)-(36)=-27$

Then factorise it, the first three terms form a perfect square

${(x + 6)}^{2} - 36 = - 27$ $(x+6)^2-36=-27$

Simplify left over numbers

${(x + 6)}^{2} = 9$ $(x+6)^2=9$

Now we solve for $x$ $x$

Square root both sides to get rid of the power

$(x + 6) = \pm 3$ $(x+6)=+-3$
$x = \pm 3 - 6$ $x=+-3-6$
$x = 3 - 6$ $x=3-6$ or $x = - 3 - 6$ $x=-3-6$
$x = - 3$ $x=-3$ or $x = - 9$ $x=-9$

How do I solve x^2 + 12x = -27x2+12x=−27x^2 + 12x = -27 by completing the square?