How do you solve using the completing the square method x^2 + 10x + 14 = -7?

Jul 8, 2016

See below.

Explanation:

The first thing you'll want to do is take the constant terms and put them to one side of the equation. In this case, that means subtracting $14$ from both sides:
${x}^{2} + 10 x = - 7 - 14$
$\to {x}^{2} + 10 x = - 21$

Now you want to take half of the $x$ term, square it, and add it to both sides. That means taking half of ten, which is $5$, squaring it, which makes $25$, and adding it to both sides:
${x}^{2} + 10 x + {\left(\frac{10}{2}\right)}^{2} = - 21 + {\left(\frac{10}{2}\right)}^{2}$
$\to {x}^{2} + 10 x + 25 = - 21 + 25$

Note that the left side of this equation is a perfect square: it factors into ${\left(x + 5\right)}^{2}$ (that's why they call it "completing the square"):
${\left(x + 5\right)}^{2} = - 21 + 25$
$\to {\left(x + 5\right)}^{2} = 4$

We can take the square root of both sides:
$x + 5 = \pm \sqrt{4}$
$\to x + 5 = \pm 2$

And subtract $5$ from both sides:
$x = \pm 2 - 5$
$\to x = + 2 - 5 = - 3$ and $x = - 2 - 5 = - 7$

Our solutions are therefore $x = - 3$ and $x = - 7$.