How do you solve the equation #x^2+6x+13=0# by completing the square?

Question

Liam Brennan · Accepted Answer

#x = -3+-2i#

The squares identity difference can be expressed as follows:

#a^2-b^2 = (a-b)(a+b)#

Use this with #a=(x+3)# and #b=2i# as follows:

#0 = x^2+6x+13#

#color(white)(0) = x^2+6x+9+4#

#color(white)(0) = (x+3)^2-(2i)^2#

#color(white)(0) = ((x+3)-2i)((x+3)+2i)#

#color(white)(0) = (x+3-2i)(x+3+2i)#

Hence:

#x = -3+-2i#

Elijah Allen · Answer

undefined To solve the equation $ x^2 + 6x + 13 = 0 $ by completing the square, follow these steps:

1. Move the constant term to the other side of the equation:
$$ x^2 + 6x = -13 $$

2. To complete the square, take half of the coefficient of $ x $ (which is 6), square it, and add it to both sides of the equation:
$$ x^2 + 6x + (6/2)^2 = -13 + (6/2)^2 $$
$$ x^2 + 6x + 9 = -13 + 9 $$

3. Simplify both sides of the equation:
$$ x^2 + 6x + 9 = -4 $$

4. Rewrite the left side of the equation as a perfect square trinomial:
$$ (x + 3)^2 = -4 $$

5. Take the square root of both sides of the equation:
$$ x + 3 = \pm \sqrt{-4} $$

6. Since the square root of a negative number is imaginary, there are no real solutions for $ x $.

Therefore, the equation $ x^2 + 6x + 13 = 0 $ does not have real solutions.