Solving radicals in an equation

When a variable sits under a root, the plan is to isolate the radical and then undo it.
You undo a square root by squaring both sides, and a cube root by cubing both sides.

First get the radical alone.
From $24 = 6\sqrt{3x}$, divide both sides by $6$ to get $4 = \sqrt{3x}$.
The root must be by itself before you raise to a power, or the other terms get in the way.

Then raise both sides to the power that cancels the root.
Squaring $4 = \sqrt{3x}$ gives $16 = 3x$, so $x = \frac{16}{3}$.
A cube root needs cubing: $\sqrt[3]{\frac{y}{x^2}} = x$ becomes $\frac{y}{x^2} = x^3$, giving $y = x^5$.

Because squaring can create false solutions, check each answer in the original.
Substitute it back and confirm both sides match and that no root is being asked to equal a negative number.
Keep only the values that truly satisfy the original equation.

Solve $2\sqrt{x} - 3 = 5$.
First isolate the radical: add $3$ to both sides to get $2\sqrt{x} = 8$, then divide by $2$ to get $\sqrt{x} = 4$.
Now square both sides: $x = 16$, and checking, $2\sqrt{16} - 3 = 8 - 3 = 5$, so it is valid.

Solve $\sqrt{4x + 1} - \sqrt{x + 4} = 0$, which has a radical on each side.
Isolate the radicals by moving one to the other side: $\sqrt{4x + 1} = \sqrt{x + 4}$.
Squaring both sides gives $4x + 1 = x + 4$, so $3x = 3$ and $x = 1$; checking, $\sqrt{5} - \sqrt{5} = 0$, so it is valid.

Solve $\sqrt[3]{x - 1} + 4 = 6$.
First isolate the radical: subtract $4$ from both sides to get $\sqrt[3]{x - 1} = 2$.
Now cube both sides: $x - 1 = 8$, so $x = 9$, and checking, $\sqrt[3]{8} + 4 = 2 + 4 = 6$.