How do you solve $c^{2} + 6 c - 27 = 0$ $c^2+6c-27=0$ by completing the square?

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How do you solve $c^{2} + 6 c - 27 = 0$ $c^2+6c-27=0$ by completing the square?

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Answer 1 · 2015-05-07T21:07:41

The answer is $c = 3$ $c=3$ , $c = - 9$ $c=-9$ .

Solve $c^{2} + 6 c - 27 = 0$ $c^2+6c-27=0$ by completing the square.

Add $27$ $27$ to both sides.

$c^{2} + 6 c = 27$ $c^2+6c=27$

Halve the coefficient of $6 c$ $6c$ then square it.
$\frac{6}{2} = 3$ $6/2=3$

$3^{2} = 9$ $3^2=9$

Add $9$ $9$ to both sides of the equation.

$c^{2} + 6 c + 9 = 27 + 9$ $c^2+6c+9=27+9$ =

$c^{2} + 6 c + 9 = 36$ $c^2+6c+9=36$

The lefthand side of the equation is now a perfect trinomial square. Factor the perfect trinomial square. $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ $a^2+2ab+b^2=(a+b)^2$

$a = c$ $a=c$ , $b = 3$ $b=3$

${(c + 3)}^{2} = 36$ $(c+3)^2=36$

Take the square root of both sides and solve for $c$ $c$ .

$c + 3 = \pm \sqrt{36}$ $c+3=+-sqrt36$

$c + 3 = \pm 6$ $c+3=+-6$

When $c + 3 = 6$ $c+3=6$ :

$c = 6 - 3$ $c=6-3$ , $c = 3$ $c=3$

When $c + 3 = - 6$ $c+3=-6$ :

$c = - 6 - 3$ $c=-6-3$ , $c = - 9$ $c=-9$

Check:

$3^{2} + 6 \cdot 3 - 27 = 0$ $3^2+6*3-27=0$ =

$9 + 18 - 27 = 0$ $9+18-27=0$ =

$27 - 27 = 0$ $27-27=0$

$- 9^{2} + 6 (- 9) - 27 = 0$ $-9^2+6(-9)-27=0$ =

$81 - 54 - 27 = 0$ $81-54-27=0$ =

$81 - 81 = 0$ $81-81=0$

Reference: http://www.regentsprep.org/regents/math/algtrig/ate12/completesqlesson.htm

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How do you solve $c^{2} + 6 c - 27 = 0$ $c^2+6c-27=0$ by completing the square?

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How do you solve c2+6c−27=0 c^2+6c-27=0 by completing the square?

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How do you solve $c^{2} + 6 c - 27 = 0$ $c^2+6c-27=0$ by completing the square?