Adding radicals

The most important warning first: a square root does not split over addition.
So $\sqrt{a} + \sqrt{b}$ is not the same as $\sqrt{a + b}$.
To see it, $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$, but $\sqrt{9 + 16} = \sqrt{25} = 5$.

You can add radicals when they are like radicals — they have the same number under the root.
Treat the root like a variable: $2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3}$, exactly the way $2x + 5x = 7x$.
Unlike radicals, such as $\sqrt{3} + \sqrt{5}$, cannot be combined into one term.

Sometimes radicals only look unlike until you simplify them.
For example, $\sqrt{12}$ simplifies to $2\sqrt{3}$, because $12 = 4 \times 3$ and $\sqrt{4} = 2$.
So $\sqrt{12} + \sqrt{3} = 2\sqrt{3} + \sqrt{3} = 3\sqrt{3}$.

In a radical equation, it helps to treat the whole root as a single quantity.
For $6 - 3\sqrt{m} = 5\sqrt{m} + 2$, think of $\sqrt{m}$ as one thing and collect it on one side.
Subtracting $5\sqrt{m}$ and $6$ from both sides gives $-8\sqrt{m} = -4$, so $\sqrt{m} = \frac{1}{2}$ and $m = \frac{1}{4}$.

Simplify $3\sqrt{5} + 4\sqrt{5}$.
These are like radicals, both with $\sqrt{5}$, so add the numbers in front.
So $3\sqrt{5} + 4\sqrt{5} = 7\sqrt{5}$.

Simplify $\sqrt{8} + \sqrt{2}$.
First simplify $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$.
Now the terms are like radicals: $2\sqrt{2} + \sqrt{2} = 3\sqrt{2}$.

Solve $4\sqrt{x} = \sqrt{x} + 9$.
Treat $\sqrt{x}$ as a single quantity and collect it: subtract $\sqrt{x}$ from both sides to get $3\sqrt{x} = 9$.
So $\sqrt{x} = 3$, which gives $x = 9$.