# Complete the square then solve the equation : x^2+2x+7=0

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We have the equation x^2+2x+7=0.

x^2+2x+7=0

=> x^2 + 2x + 1 + 6 = 0

=> (x + 1)^2 + 6 = 0

=> (x + 1)^2 = -6

=> x + 1 = - i* sqrt 6 and x +1 = i*sqrt 6

=> x = -1 + i*sqrt 6 and x = -1 - i*sqrt 6

**Therefore x is -1 - i*sqrt 6 and -1 + i*sqrt 6**

Complete the square then solve the equation: x^2+2x+7=0

x^2+2x+7 = 0.

=> x^2+2x = -7.

The left side x^2+2x becomes a perfect square if we add 1. x^2+2x+1 = (x+1)^2. So we add 1 to both sides of the equation:

(x+1)^2 = -7+1 = -6.

=> (x+1)^2 = -6

We take the square root:

x+1 = (-6)^(1/2), Or x+1 = - (-6)^(1/2).

=> **x= (-6)^(1/2)-1, or x = -(-6)^(1/2)-1.**

For the beginning, we'll subtract 7 both sides, to move the constant on the right side of the equation.

x^2 + 2x = -7

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

x^2 +2x + 1 = -7 + 1

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

(x + 1)^2 = -6

x + 1 = sqrt -6

Since sqrt -1 = i, we'll get:

x + 1 = isqrt 6

We'll subtract 1 both sides:

x1 = -1 + isqrt 6

x2 = -1 - isqrt 6

**The complex solutions of the equation are: {-1 + isqrt 6; -1 - isqrt 6}.**