Say that **you've only just learned about roots and how to simplify radicals**. That doesn't stop your teacher from giving the whole class a task to **try to do on your own** while they sit behind their desk with a phone in hand. They seem to think simplifying radical expressions will take you long enough for them to browse through the newsfeed. **Why don't we prove them wrong?**

The task is to **find the sum, product, and quotient of** **2√6** **and** ** ^{4}√64**. In other words, we need to compute

**2√6 +**,

^{4}√64**2√6 ×**, and

^{4}√64**2√6 /**.

^{4}√64Well, **how fortunate that we have Omni's simplifying radicals calculator** at hand!

Let's start with **the sum**. In the calculator, we see an option to **choose the radical expression that we want to find**. We're interested in adding two values, so we choose "*sum*" under "*Expression*." That will trigger a symbolic representation of the operation to appear.

next, we need to **input the numbers**. The simplify radical expressions calculator shows the sum as **a × ^{n}√b + c × ^{m}√d**, so for our case, we input

**a = 2**, **b = 6**, **n = 2**, **c = 1**, **d = 64**, **m = 4**.

(note how **n = 2** is the default since, usually, we deal with square roots. Also, we didn't really need to input **c = 1** – the calculator understands blank fields for **a** and **c** as **1**s.)

Once we give all the numbers, we can **read off the result from underneath** the variable fields. Observe how the calculator also gives **a step-by-step solution** to your problem.

For **the product and quotient**, we repeat the above steps (the **a**, **b**, and so on don't change), but choose the correct option under "*Expression*" – * product* and

*, respectively.*

**quotient**nevertheless, for the horrific times when you can't use the internet, let's see **how to simplify the radicals ourselves** without our tool's help.

We begin with **the sum**: **2√6 + ^{4}√64**. First of all, we need to

**find prime factorizations**of the two numbers under the radicals:

**6 = 2 × 3**,

**64 = 2 × 2 × 2 × 2 × 2 × 2 = 2 ^{6}**.

The first root is of order **2**, so we need to find pairs of the same number in the factorization. We see that **there is none**, so that summand is already in its simplest form.

For the second, **we need fours**. Indeed, there is one such solution (four **2**s), which leaves two **2**s alone. We pull the numbers representing the groups of four out of the radical and keep the rest inside.

**2√6 + ^{4}√64 = 2√6 + ^{4}√(2^{6}) = 2√6 + 2 × ^{4}√(2^{2})**

Unfortunately, **this is not over yet**. In the second summand, we see that the order of the root and the powers of all (i.e., of the only one) numbers under it **have a common factor** – **2**. Therefore, we reduce the two numbers by that factor.

**2√6 + ^{4}√64 = 2√6 + 2 × ^{4}√(2^{2}) = 2√6 + 2√2**

Note how we didn't write the ** ^{2}** with the second radical because, by convention,

**we write square roots without that number**.

The two summands have different numbers under the radicals (but the same root orders), so we can't add them – **this is the simplest form of the expression**.

Now for simplifying **the radical expression with the product**: **2√6 × ^{4}√64**. The two roots have orders

**2**and

**4**, respectively, and

**lcm(2,4) = 4**. We follow the instructions given in the above section and get:

**2√6 × ^{4}√64 = 2 × ^{4}√(6^{2} × 64) = 2 × ^{4}√2304**.

Next, **we find the prime factorization** of the number under the root:

**2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 = 2 ^{8} × 3^{2}**.

We look for **fours of the same prime** in the factorization and find two: a couple of fours of **2**s. We pull these out of the radical and get

**2√6 × ^{4}√64 = 2 × ^{4}√2304 = 2 × ^{4}√(2^{8} × 3^{2}) = 2 × 2 × 2 × ^{4}√3^{2} = 8 × ^{4}√3^{2}**.

We have a similar situation to what we had with the sum: **the order of the root and the primes' powers under it have a common factor**. We reduce them and get our final answer:

**2√6 × ^{4}√64 = 8 × ^{4}√3^{2} = 8√3**.

Lastly, let's see how to simplify **the radical expression with** the quotient: **2√6 / ^{4}√64**. We recall that

**lcm(2,4) = 4**and the instructions from the above section to obtain:

**2√6 / ^{4}√64 = (2 / 64) × ^{4}√(6^{2} × 64^{3}) = 0.03125 × ^{4}√9,437,184**.

**We find the prime factorization** of the number under the root:

**9,437,184 = 2 ^{20} × 3^{2}**,

and look for **fours of the same primes**. In this case, we have five fours of **2**. We pull these out of the radical and get:

**2√6 / ^{4}√64 = 0.03125 × ^{4}√9,437,184 = 0.03125 × ^{4}√(2^{20} × 3^{2}) = 0.03125 × 2^{5} × ^{4}√(3^{2}) = ^{4}√(3^{2})**.

Again, we can **reduce the order of the root and the powers of the primes** under it. That gives us a final answer of:

^{2}√6 / ^{4}√64 = ^{4}√(3^{2}) = √3

Arguably, it took some time to write all the details, but the operations themselves weren't too terrible. Still, it makes us appreciate **how much work the simplifying radicals calculator can save us**. And after reading through this article, even doing it by hand must take less than the teacher suspects!