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

2 Answers
Jan 28, 2016

Answer:

# x = ± sqrt3 - 3 #

Explanation:

By adding 9 to both sides of the equation to obtain:

#( x^2 + 3x + 9 )+ 6 = 9#

now #(x + 3 )^ 2 = 9-6 = 3#

# ( x + 3 )^2 color(black)( " is a perfect square ") #

Taking the 'square root' of both sides :

# sqrt((x+ 3 )^2 )= sqrt3#

hence x + 3 = ± #sqrt3 #

so x = ±#sqrt3 - 3 #

Answer:

there are two #x# values

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

#x_2=(-3-sqrt(15)i)/2#

Explanation:

Completing the square method:
Do this only when the numerical coefficient of #x^2# is #1#.
Start with the numerical coefficient of #x# which is the number #3#.
Divide this number by 2 then square the result. That is

#(3/2)^2=9/4#

Add #9/4# to both sides of the equation

#x^2+3x+9/4+6=0+9/4#

the first three terms now become one group which is a PST-Perfect Square Trinomial

#(x^2+3x+9/4)+6=9/4#

#(x+3/2)^2+6=9/4#

#(x+3/2)^2=9/4-6# after transposing the #6# to the right side

#(x+3/2)^2=(9-24)/4#

#sqrt((x+3/2)^2)=+-sqrt((9-24)/4)#

#x+3/2=+-sqrt((-15)/4)#

#x+3/2=+-sqrt(-15)/sqrt(4)#

#x+3/2=+-sqrt(-15)/2#

Finally, transpose the #3/2# to the right side of the equation

#x=-3/2+-sqrt(-15)/2#

take note: #sqrt(-15)=sqrt(15)*sqrt(-1)=sqrt(15)i#

therefore

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

there are two #x# values

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

#x_2=(-3-sqrt(15)i)/2#