SOLUTION: Here are a few problems that I am not sure if I got them right. Thanks. Paul 1) problem: square root of 3x + 9, - 12 = 0 square root of 3x + 9 = 12 (square root of 3x +

Question 163508: Here are a few problems that I am not sure if I got them right. Thanks. Paul
1) problem: square root of 3x + 9, - 12 = 0
square root of 3x + 9 = 12
(square root of 3x + 9)^2 = (12)^2
3x + 9 = 144
-9 -9
3x = 135
3x/3 = 135/3
answer: x = 45

2) problem: square root of 3x, - square root of 3x-5, = 1
1 - , square root of 3x = square root of 3x-5
(1 - square root of 3x)^2 = (square root of 3x-5)^2
1 - 2 square root of 3x - 3x = -3x-5
6 = 2 square root of 3x
(4)^2 = (square root of 3x)^2
16/3 = 3x/3
answer: x = 16/3 (I think this answer is 3 but not sure on how to get it)

3) problem: ^3 square root of x - 2, = 3
(^3 square root of x -2 ) = (3)^3
x - 2 = 27
answer: x = 29

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Here are a few problems that I am not sure if I got them right. Thanks. Paul
1) problem: square root of 3x + 9, - 12 = 0
square root of 3x + 9 = 12
(square root of 3x + 9)^2 = (12)^2
3x + 9 = 144
-9 -9
3x = 135
3x/3 = 135/3
answer: x = 45
:
This is right, check it yourself, substitute 45 for x in the original equation
:
2) problem: square root of 3x, - square root of 3x-5, = 1
1 - , square root of 3x = square root of 3x-5
(1 - square root of 3x)^2 = (square root of 3x-5)^2
Here you should have:
1 - 2*sqrt(3x) + 3x = 3x - 5
Combine on the right, leaving the radical on the left
-2*Sqrt(3x) = 3x - 3x - 5 - 1
-2*sqrt(3x) = -6
divide both sides by -2
sqrt(3x) = +3
Square both sides
3x = 9
x = 3; here again you can confirm this, substitute 3 for x in the original equation
:
:
3) problem: ^3 square root of x - 2, = 3
(^3 square root of x -2 ) = (3)^3
x - 2 = 27
answer: x = 29
:
This is right, you can write it: CubeRt(x-2) = 3
Substitution proves this one too
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