Question

Add and subtractin radical √2+√8

Answer 1

To simplify the expression \(\sqrt{2} + \sqrt{8}\), we can start by simplifying \(\sqrt{8}\):

\[
\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \cdot \sqrt{2} = 2\sqrt{2}
\]

Now, we can rewrite the original expression:

\[
\sqrt{2} + \sqrt{8} = \sqrt{2} + 2\sqrt{2}
\]

Next, we can combine the like terms:

\[
\sqrt{2} + 2\sqrt{2} = (1 + 2)\sqrt{2} = 3\sqrt{2}
\]

So,

\[
\sqrt{2} + \sqrt{8} = 3\sqrt{2}
\]

Now, if you want to see the subtraction as well, let's say we want to find \(\sqrt{2} - \sqrt{8}\):

Using the simplified form of \(\sqrt{8}\):

\[
\sqrt{2} - \sqrt{8} = \sqrt{2} - 2\sqrt{2} = (1 - 2)\sqrt{2} = -\sqrt{2}
\]

So the results are:
- Addition: \(\sqrt{2} + \sqrt{8} = 3\sqrt{2}\)
- Subtraction: \(\sqrt{2} - \sqrt{8} = -\sqrt{2}\)