Simplifying Roots

A square root of any number, for example √x, is actually a number that when multiplied by itself gives us $X$.
For example:
$\sqrt{25}$
$5\cdot5=25$
Therefore:
$\sqrt{25}=5$

Another example:
$\sqrt{36}=6$
What will happen if we're asked what is the square root of $72$ for example:
$\sqrt{72}=?$
This is exactly where the square root simplification technique comes in handy!

This technique simplifies the number inside the root by factoring it, making it much easier for us to find the solution!
We can express $72$ like this: $36\cdot2$
And actually write
$\sqrt{72}=\sqrt{2\cdot36}$
We can write this exercise like this according to the laws of exponents:
$\sqrt{2\cdot36}=\sqrt{36}\cdot\sqrt2$
Great! We know what the square root of $36$ is, so we can write it in this form:
$\sqrt{72}=6\cdot\sqrt2$

Let's summarize the solution steps:

1. Check if the number under the root can be broken down into simple factors that can easily be square rooted.
2. For numbers that can be square rooted, we extract the root, and for those that cannot, we leave them with the root.
3. Verify that there are only multiplication operations between the factors.

Tip!
How do we know which factors are best to break down the number inside the square root?
We will always look for the largest factor that has a whole square root!
For example:
$\sqrt{50}=$
Let's think about which is the largest factor we can take out that has a square root?
The answer is $25$. If we take out the factor $25$ we obtain the following:
$\sqrt{50}=\sqrt{25}\cdot\sqrt2$
We can then proceed to write that:
$\sqrt{50}=5\sqrt2$

If we were to factor out numbers like $5$ and $10$ or $50$ and $1$ we wouldn't be able to progress with solving the exercise.

Now let's move on to more complex exercises!

Here is the exercise $5\sqrt{27}=$
Don't be intimidated by its scary appearance! Just imagine a multiplication operation between $5$ and the square root sign, and continue to factor the number $27$ as you learned before.
The largest number in $27$ that we can find a square root for is $9$.
$27:9=3$
Therefore we'll write the following:
$5\sqrt{27}=5\sqrt9\cdot\sqrt3$
We know that $\sqrt9=3$ hence we can substitute $3$ for $\sqrt9$ and obtain:
$5\sqrt{27}=5\cdot3\cdot\sqrt3$
$5\sqrt{27}=15\sqrt3$

Another exercise:
$2\sqrt{72}=$
Think about the largest factor that can be extracted from $72$ and apply a square root to it..
The answer is of course $9$!
$72:9=8$
Therefore we can write
$2\sqrt{72}=$
$2\sqrt9\cdot\sqrt8$

Pay attention!!
We know that $\sqrt9$ is $3$ therefore we can write the exercise like this:
$2\cdot3\cdot\sqrt8$

But what about $\sqrt8$?
We can also factor $\sqrt8$!
The largest factor we can take out of $8$ and apply a square root to is $8$.
$8:4=2$
So the two factors we'll take out are 2 and 4.
We observe that:
$2\sqrt{72}=2\cdot3\cdot\sqrt4\cdot\sqrt2$
We know that
$2\sqrt{72}=2\cdot3\cdot\sqrt4\cdot\sqrt2$
Therefore, we'll insert the data into the exercise and obtain:
$2\sqrt{72}=2\cdot3\cdot2\cdot \sqrt2$
$2\sqrt{72}=12\sqrt2$

Keep practicing and make the exam easier for yourself!
$6\sqrt{25}=$

Note- we won't always rush to find and extract factors on autopilot. We know that $\sqrt{25}=5$
therefore this exercise, for example, is really quite simple. We obtain the following solution:
$6\sqrt{25}=6\cdot5=30$

Another exercise:
$5\sqrt{29}=$
Let's identify the largest factor we can take out of $29$ that has a whole square root?
The answer is that there isn't one. The factors of $29$ are only $29$ and $1$
Therefore we cannot simplify this square root.

What about this exercise?
$10\sqrt{360}=$
With this large number, we'll work in steps.
Let's look at $360$. From the number $36$ we can extract $9$ hence we'll write as follows:
$10\sqrt{360}=10\cdot\sqrt9\cdot\sqrt{40}$
From $40$ we can extract multiples of $10$ and $4$
We obtain the following solution as seen below:
$10\sqrt{360}=10\cdot\sqrt9\cdot\sqrt{10}\cdot\sqrt4$
Now we'll solve what we can and get:
$10\sqrt{360}=10\cdot3\cdot2\cdot\sqrt{10}$
$10\sqrt{360}=60\sqrt{10}$