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

1 Answer
Jul 8, 2016

Answer:

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+10x=-7-14#
#->x^2+10x=-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+10x+(10/2)^2=-21+(10/2)^2#
#->x^2+10x+25=-21+25#

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

We can take the square root of both sides:
#x+5=+-sqrt(4)#
#->x+5=+-2#

And subtract #5# from both sides:
#x=+-2-5#
#->x=+2-5=-3# and #x=-2-5=-7#

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