# How to determine the solutions of equation x^2+2x-8=0, completing the square?

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We have the equation x^2 + 2x - 8 = 0 to solve by completion of square.

We know that (a + b)^2 = a^2 + b^2 + 2ab

x^2 + 2x - 8 = 0

=> x^2 + 2x + 1 - 8 -1 = 0

=> (x + 1)^2 - 9 = 0

=> (x + 1)^2 = 9

Now we get

x + 1 = sqrt 9 and x + 1 = -sqrt 9

x = -1 + sqrt 9 and x = -1 - sqrt 9

x = -1 + 3 and x = -1 - 3

x = 2 and x = -4

**The solution of the equation is x = 2 and x = -4**

x^2+2x=8

x^2+2x+(2/2)^2=8+(2/2)^2

(x+1)^2=9

x+1= +3 and -3

x=2 or -4

to complete the square, bring the number which is c in the general quadratic equation ax^2+bx+c=0 to the other side of the equation.

divide b in bx by 2 and square the number. add this to both sides of the equation. Remember what you do to the left, you do to the right.

therefore add whatever the number of b/2 to x to have (x+b/2)^2

equate whatever is on the other side and solve!

For the beginning, we'll add 8 both sides, to move the constant on the right side of the equation, so that being more clear what we have to add to the left side to complete the square.

x^2 +2x = 8

We'll complete the square by adding the number 1 to both side, to get a perfect square to the left side.

x^2 +2x + 1 = 8 + 1

We'll write the left side as a perfect square:

(x + 1)^2 = 9

x + 1 = sqrt 9

x + 1 = 3

x1 = 3 - 1

x1 = 2

x2 =-3 - 1

x2 = -4

**The real solutions of the equation are: { -4; 2}.**