x^2+12x+31=0

Question

solve by completing the square, please.

Answer 1

• x²+12x+31=0   (Original problem)
• x²+12x=-31      (Move constant to right)
• x²+12x+36=-31+36 (add the 36 to make the left side a perfect square, and add it to the right to maintain balance)
• (x+6)²=5     (Factor the perfect square that you created...this should be easy, because you had to think of this for the last step)
• x+6= +/- √5   (Take the square root of both sides)
• x= -6 +/- √5

Answer 2

x²+12x+31=0

x²+12x      =-31

12/2=6, 6²=36

x²+12x+36=36-31

(x+6)²=5

take the square root of both sides

x+6=+√5

x=-6+√5

x=-6-√5

x=-6+√5

 

Answer 3

First, move the constant term, 31, to the other side of the equation:
x^2 + 12x = -31.
Next, you need to look at the coefficient in front of the x, in this case, it's 12. You take half of that, which is 6, and square it, which is 36. Now you have to add this number to both sides:
x^2 + 12x + 36 = -31 + 36.
Simplify the right side of the equation:
x^2 + 12x + 36 = 5.
Now the left side of the equation is a perfect square, so you can factor it:
(x + 6)^2 = 5.
From here you can solve this for x. Take square roots of both sides, and then subtract 6 from both sides:
sqrt((x + 6)^2) = +/- sqrt(5)
x + 6 = +/- sqrt(5)
x = -6 + sqrt(5) and x = -6 - sqrt(5) are your two solutions. Hope this helps!