How do you solve $root3x -4 = 2$ ?

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Answer 1 · 2016-05-02T11:00:55

$x=216$

Add 4 to both sides

$root(3)(x)+0=2+4$

$root(3)(x)=6$

Example: $(sqrt(4))^2 = 4$ In squaring the square root I have reversed the process of square rooting 4

Using the above method

Cube both sides

$x=6^3$

$x=216$

How do you solve root3x -4 = 2root3x -4 = 2?