Question

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

Answer 1

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

Solve `c^2+6c-27=0` by completing the square.

Add `27` to both sides.

`c^2+6c=27`

Halve the coefficient of `6c` then square it.
`6/2=3`

`3^2=9`

Add `9` to both sides of the equation.

`c^2+6c+9=27+9` =

`c^2+6c+9=36`

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

`a=c`, `b=3`

`(c+3)^2=36`

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

`c+3=+-sqrt36`

`c+3=+-6`

When `c+3=6`:

`c=6-3`, `c=3`

When `c+3=-6`:

`c=-6-3`, `c=-9`

Check:

`3^2+6*3-27=0` =

`9+18-27=0` =

`27-27=0`

`-9^2+6(-9)-27=0` =

`81-54-27=0` =

`81-81=0`

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