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

Question

Eva Abramson · Accepted Answer

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

Before we finish the square, let's rewrite our equation:

#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#

Taking both sides' square roots yields

#(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#

Eva Adamson · Answer

undefined To solve the equation by completing the square, follow these steps:

1. Move the constant term to the other side of the equation:
   x^2 + 2x = 8

2. To complete the square, take half of the coefficient of x (which is 1) and square it. Add this value to both sides of the equation:
   x^2 + 2x + (2/2)^2 = 8 + (2/2)^2
   x^2 + 2x + 1 = 8 + 1
   x^2 + 2x + 1 = 9

3. Rewrite the left side as a perfect square trinomial and simplify the right side:
   (x + 1)^2 = 9

4. Take the square root of both sides:
   x + 1 = ± √9

5. Simplify the right side:
   x + 1 = ± 3

6. Solve for x by subtracting 1 from both sides:
   x = -1 ± 3

7. This results in two solutions:
   x = -1 + 3 = 2
   x = -1 - 3 = -4

So, the solutions to the equation x^2 + 2x - 8 = 0 by completing the square are x = 2 and x = -4.