# How do you solve (x+7)^2= -12?

May 22, 2016

No solutions among real numbers.
Among complex numbers the two solutions are
${x}_{1} = - 7 + 2 \sqrt{3} i$ and
${x}_{2} = - 7 - 2 \sqrt{3} i$

#### Explanation:

When solving equations it is very important to define the domain of its solutions.
Are we looking for Integer solution? Rational solutions? Real solutions? Complex solutions?

Traditionally, most of the equations addressed in high school Algebra are to be solved in real numbers. This problem does not specify the domain, so we will consider two cases: real_solutions and _complex solutions.

There is no real $x$ that solve this equation because on the left we will always have a non-negative number, no matter which real $x$ we substitute since it's a square ${\left(x + 7\right)}^{2}$, while on the right we have negative number $- 12$.

If we expand our search for solutions to complex numbers, and use the classic complex number $i$ that, being squared, gives $- 1$, we can extract the square root from both sides of the equation getting
$x + 7 = \pm \sqrt{- 12}$ or
$x + 7 = \pm \sqrt{- 1} \sqrt{12}$ or
$x + 7 = \pm i \cdot \sqrt{12}$ or
$x + 7 = \pm i \cdot 2 \sqrt{3}$
which gives two solutions for $x$:
${x}_{1} = - 7 + 2 \sqrt{3} i$ and
${x}_{2} = - 7 - 2 \sqrt{3} i$

CHECK (never skip this step when solving equations!!!)
${\left({x}_{1} + 7\right)}^{2} = {\left(2 \sqrt{3} i\right)}^{2} = 12 \cdot {i}^{2} = 12 \cdot \left(- 1\right) = - 12$
${\left({x}_{2} + 7\right)}^{2} = {\left(- 2 \sqrt{3} i\right)}^{2} = 12 \cdot {i}^{2} = 12 \cdot \left(- 1\right) = - 12$