1. I know the solutions for this are 5 and 8, but I can't remember the steps for it: http://gyazo.com/7082adc73559e8f065ebc1d35838d6a9
2. I don't understand how the domain for (cuberoot(z^2))^-2) is 1/(z(cuberoot(z))

Question

1. I know the solutions for this are 5 and 8, but I can't remember the steps for it: http://gyazo.com/7082adc73559e8f065ebc1d35838d6a9
2. I don't understand how the domain for (cuberoot(z^2))^-2) is 1/(z(cuberoot(z))

anonymous · Answer

sorry omg they're two different problems I was asking about. :)

anonymous · Answer

the second one looks like this :) http://gyazo.com/c54749e031d93018c26f48c179003571

hartnn · Answer

for 1st, 
square both sides, what do u get ?

anonymous · Answer

do you set it up like (√(3x+1))(√(3x+1))=(3+√(x-4))(3+√(x-4))

anonymous · Answer

wait when you square each side do you just get 3x+1=9+(x-4)

anonymous · Answer

because when I do that, I get like 2x-4 or something and I don't think that's right because aren't you supposed to come out with a factored equation?

hartnn · Answer

when you square both sides, u get
$3x+1 = (3+\sqrt{x-4})^2$

hartnn · Answer

$3x+1 = 3^2 + 2\times 3 \times \sqrt {x-4} + (x-4)$
got this step ?

anonymous · Answer

where do you get the 2 and 3 from in the second equation? because squaring the second part of the equation is a little confusing

hartnn · Answer

$(a+b)^2 = a^2+2ab+b^2$
i applied thiso n right side!