How do you solve $3x^2+12x+81=15$ by completing the square?

Question

3128 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-30T09:25:45

$x = -2+-i sqrt(22)$

Note that all of the terms are divisible by $3$ , so let's divide both sides of the equation by $3$ to get:

$x^2+4x+27=5$

To make the left hand side into a perfect square, subtract $23$ from both sides to get:

$x^2+4x+4 = -22$

The left hand side is now $(x+2)^2$ , so our equation is:

$(x+2)^2 = -22$

The square of any real number is non-negative, so this quadratic equation only has non-real complex solutions.

If you want to proceed further, note that if $n > 0$ then:

$sqrt(-n) = i sqrt(n)$

where $i$ is the imaginary unit, satisfying $i^2 = -1$ .

So we find:

$(x+2)^2 = (i sqrt(22))^2$

and hence:

$x+2 = +-i sqrt(22)$

Subtracting $2$ from both sides we get:

$x = -2+-i sqrt(22)$

How do you solve 3x^2+12x+81=153x^2+12x+81=15 by completing the square?