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Could you please check my answers and help with the ones I can't figure out-
1. Solution to the radical function sqrt(1/2x+1) = 1 is x = 0
True or False I think it is true according to my calculations

2.Solve x = sqrt(8x)
8x = x^2
x^2 - 8x = 0
x(x-8) = 0
x = 0
x-8 =0
x=8
the only solution is x = 0, correct or nobecause x = 8 doesn't work-
3. Simply (3x^2y^-2)^-2/9x^-5y
a.xy^3
b. xy^3/3
xy^3/9
d. xy^3/81

I think it is xy^3/81 but I'm not sure of my calculations

4. Solve 4sqrt(6x-2)>4
a. 1/3<=x<43
b. 1/3<x<43
x>=43
x>43
I think x>43
Thank you

• Algebra-Please check my calculations -

1. true if the equation is
sqrt((1/2)x+1) = 1, or
sqrt(x/2+1) = 1

2. x=√(8x)
"the only solution is x = 0, correct or nobecause x = 8 doesn't work- " ???
x=0 works: 0=√(8*0)
x=8 also works √(8*8)=8

3. Correct.

4. Solve 4sqrt(6x-2)>4
Divide by 4 to get
sqrt(6x-2)>1
For this to happen, 6x-2>1 or x>(1/2)
I suspect there is a misprint or misinterpretation of the question.

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