How do you solve the equation by completing the square: #x^2 + 2x - 8 = 0#?

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Answer 1 · 2016-07-12T00:21:58

#x = 2# and #x = -4#

Let's rewrite our equation first, before completing the square:

#x^2+2x-8=0#

#x^2+2x + ? = 8 + ?#

To complete the square, we take the coefficient on the #x#-term, namely #2#, divide it by #2#, and square the result, giving us

#(2/2)^(2) = 1^2 = 1#

By replacing the #?# marks with our number #1#, we get

#x^2+2x+1 = 9#

In this case, we are looking for two numbers whose product gives us #1# and when added together, gives us #2#.

Since #1 * 1 = 1# and #1+1 = 2#, we can see that the numbers are #1# and #1#, which means that we can rewrite our equation in the following way:

#(x+1)^(2) = 9#

By taking the square root of both sides we get

#(x+1) = ± sqrt(9)#

#x+1 = ± 3#

Subtracting #1# from both sides gives us

#x = ± 3 -1#

So our solutions are #x = 2# and #x = -4#