Difficult math problem

-3x[ 1/2(x+1)^(-1/2)] - 3sqrt(x+1) = 0
Thanks for the help!

You may want to double check this, but I believe the answer is x= -2/3
1) negative powers mean the base of the exponent switches from the numerator to the denominator, so -3x[1/2(1/[x+1]^1/2)]-3sqrt(x+1)=0
2) a power of 1/2 is the same thing as a sqrt, and denominators can be combined in multiplication
-3x [1/2sqrt(x+1)]-3sqrt (x+1)=0
3) multiplication
(-3x)/(2sqrt[x+1]) -3qrt(x+1)=0
4) multiply everything by 2sqrt (x+1), which cancels out the fraction in the first term, is multiplied with the second term, and doesn't affect the other side of the equation because it's 0.
-3x -(3sqrt[x+1])(2sqrt[x+1]) = 0
5) multiply two square roots of the same number together gives that number
-3x - (6[x+1])=0
6) distributive property
-3x-6x-6=0
7) combine like terms
-9x-6=0
8) add 6 to both sides of the equation
-9x=6
9) divide both sides of the equation by -9 to isolate the variable
x= -6/9
10) simplify the fraction
x= -2/3

If you raise something to the power of a fraction that's the same as taking roots of it so (x+1)^(1/2)=sqrt(x+1).

If you raise something to the power of -1 then that's the same as taking the reciprocal
so (x+1)^(-1/2)=1/sqrt(x+1)

Now if I multiply by sqrt(x+1), I get -3x/2 - 3(x+1) = 0

If I multiply by 2 then I get:
-3x - 6x + 6 = 0
-9x = -6
x = (-6)/(-9) = 2/3

divide by (-3)

x/(2√(x + 1)) + √(x + 1) = 0

[x + 2 (√(x + 1) )^2]/(2√(x + 1) ) = 0

[x + 2 x + 2]/(2√(x + 1) ) = 0

(3x + 2)/(2√(x + 1)) = 0

3x + 2 = 0

x = -2/3


If you intended this

- 3x(1/(2(x + 1)^(- 1/2))) - 3√(x + 1) = 0

divide by (-3) all terms

x(1/(2(x + 1)^(- 1/2))) + √(x + 1) = 0

x √(x + 1)/2 + √(x + 1) = 0

√(x + 1) (x/2 + 1) = 0

x = -1

x = - 2