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

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Answer 1 · 2017-06-30T09:25:45

$x = - 2 \pm i \sqrt{22}$ $x = -2+-i sqrt(22)$

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

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

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

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

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

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

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 $n > 0$ $n > 0$ then:

$\sqrt{- n} = i \sqrt{n}$ $sqrt(-n) = i sqrt(n)$

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

So we find:

${(x + 2)}^{2} = {(i \sqrt{22})}^{2}$ $(x+2)^2 = (i sqrt(22))^2$

and hence:

$x + 2 = \pm i \sqrt{22}$ $x+2 = +-i sqrt(22)$

Subtracting $2$ $2$ from both sides we get:

$x = - 2 \pm i \sqrt{22}$ $x = -2+-i sqrt(22)$

How do you solve 3x2+12x+81=153x^2+12x+81=15 by completing the square?