How do you solve #x^2 + 3x +6 =0# by completing the square?

Question

Eva Adamson · Accepted Answer

# x = ± sqrt3 - 3 #

To get at the following, add 9 to both sides of the equation: 9# = #(x^2 + 3x + 9) + 6 Therefore, #(x + 3)^2 = 9-6 = 3#. #(x + 3)^2 color(black)("is a perfect square") # Taking both sides' "square root": #sqrt((x+3)^2 )= sqrt3# x + 3 = ± #sqrt3 #, thus x = ±#sqrt3 - 3 #, thus

Abigail Allen · Answer

there are two #x# values

#x_1=(-3+sqrt(15)i)/2#

#x_2=(-3-sqrt(15)i)/2# Squaring the result: This method should only be completed if the numerical coefficient of #x^2# is #1#. To begin, take the numerical coefficient of #x#, which is #3#. Divide this number by two and square the result. #(3/2)^2 = 9/4# On both sides of the equation, add #9/4#. #9/4+6+x^2+3x=0+9/4# the initial three terms now form a PST (Perfect Square Trinomial) group. 9/4#= #(x^2+3x+9/4)+6 #(x+3/2)^2+6=9/4# #(x+3/2)^2=9/4-6# once the #6# has been flipped to the right. #(x+3/2)^2 = (9–24)/4# (x+3/2)^2) = +-sqrt((9-24)/4)# #(-15)/4)# +-sqrt(x+3/2)# +-sqrt(-15)/sqrt(4)# #x+3/2 +-sqrt(-15)/2# = #x+3/2# Lastly, move the #3/2# to the equation's right side. #x = -3/2 + -sqrt(-15)/2# Remember that #sqrt(-15)=sqrt(15)*sqrt(-1)=sqrt(15)i#. Consequently #x = 3/2 + -(sqrt(15)i)/2# Two #x# values exist. (-3+sqrt(15)i)/2# is #x_1 (-3-sqrt(15)i)/2# is #x_2

Genesis Allen · Answer

undefined To solve the equation x^2 + 3x + 6 = 0 by completing the square, follow these steps:

1. Move the constant term to the other side:
   x^2 + 3x = -6

2. To complete the square, add (3/2)^2 = 9/4 to both sides:
   x^2 + 3x + 9/4 = -6 + 9/4
   (x + 3/2)^2 = -24/4 + 9/4
   (x + 3/2)^2 = -15/4

3. Take the square root of both sides:
   x + 3/2 = ±√(-15/4)
   x + 3/2 = ±(i√15)/2

4. Subtract 3/2 from both sides:
   x = -3/2 ± (i√15)/2

Therefore, the solutions to the equation x^2 + 3x + 6 = 0 are:
x = -3/2 + (i√15)/2
x = -3/2 - (i√15)/2