solving equations containing radicals 6+2x(3)^(1/2)=0

[bluewhale210]

iam sorry that not the it it 6+2x to the square root of three=0 but my key dosnt have a radical sign any ways i getting confused but i got the answer + or - the square root of 27 =x the problem is that the book (were iam working from) say the answer is -sqare root 3 is =x here my work
6+2x(square root)3=0
-6 -6
2x(square root)3=-6
/2x /2x
(square root)3=-6/2x
((square root)3)^2)=(-6/2x)^2
3=36/(4x^2)
(36)3=36/(4x^2)(36)
108=4x^2
108/4=(4x^2)/4
27=x^2
+or - (square root)27=x
when i check it doesnt equare 0 i know it could just be a extraneous solution but i really am not sure if you could help even just to point out a mistake i done.
thank you

 

[arthur ohlsten]

a squared = a^2
a cubed = a^3
square root a=a^1/2
cube root of a=a^1/3
a to the nth power = a^n
the nth root of a=a^1/n

you never need radical sign with this notation

6+2x[3^1/2]=0
subtract 6 from each side
2x[3^1/2]=-6
divide both sides by 2
x[3^1/2]=-3
divide both sides by 3^1/2
x=-[3/3^1/2]
x=-3^1/2 answer
Arthur

 

[bluewhale210]

thank you ever much for the help