# How do you simplify sqrt(108) ?

May 14, 2017

Well, first, let's explain how you simplify a square root:

Let's say the problem is this
$\sqrt{{x}^{2} \times {y}^{3}}$

Our first task is to expand all our components

$\sqrt{x \times x \times y \times y \times y}$

Now we need to look at our index, the $3$ in the top left corner

Whatever that number is, that's the value of the exponent we're trying to create. If it's a $\sqrt{\textcolor{w h i t e}{x}}$, we want an ${x}^{2}$. If it's a $\sqrt{\textcolor{w h i t e}{x}}$, we want a ${x}^{3}$.

So, we are looking for ${x}^{3}$ inside the root, or $x \times x \times x$.
Back to our example:

Let's condense everything we can to give us a cube root

$\sqrt{x \times x \times {y}^{3}}$

When we have a number that matches the index, we can take it out of the root:

$y \sqrt{x \times x}$

We can simplify that to $y \sqrt{\sqrt{{x}^{2}}}$

$\textcolor{w h i t e}{0}$

$\sqrt{108}$

expand

$\sqrt{2 \times 3 \times 3 \times 3 \times 2}$

what's the index?
$2$
What can we change to something squared

$\sqrt{{2}^{2} \times {3}^{2} \times 3}$

Bring out the squared

That gives us $2 \times 3 \sqrt{3}$ or $6 \sqrt{3}$

$\textcolor{w h i t e}{0}$

Let's look at the other option

$2 \sqrt{27}$($\sqrt{4 \cdot 27}$)

$2 \sqrt{4 \cdot 27}$ equals $\sqrt{108}$

$2 \sqrt{27} \cdot \times \sqrt{108}$

$2 \sqrt{3 \times 3 \times 3} \times \sqrt{2 \times 2 \times 3 \times 3 \times 3}$

$2 \sqrt{{3}^{2} \times 3} \times \sqrt{{2}^{2} \times {3}^{2} \times 3}$

$2 \times 3 \sqrt{3} \times 6 \sqrt{3}$

$36 \sqrt{3 \times 3}$

$36 \times 3$

$108$

$108$ does not equal $\sqrt{108}$, so the second option cannot be correct. Hopefully this helps!

May 17, 2017

$\sqrt{108} = 6 \sqrt{3} = 3 \sqrt{12} = 2 \sqrt{27}$

the "simplest" being $6 \sqrt{3}$.

#### Explanation:

When $a \ge 0$ and/or $b \ge 0$ then:

$\sqrt{a b} = \sqrt{a} \sqrt{b}$

Also, if $a \ge 0$ then:

$\sqrt{{a}^{2}} = a$

The prime factorisation of $108$ is:

$108 = 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3$

So we find:

$\sqrt{108} = \sqrt{{2}^{2} \cdot 27} = \sqrt{{2}^{2}} \sqrt{27} = 2 \sqrt{27}$

$\sqrt{108} = \sqrt{{3}^{2} \cdot 12} = \sqrt{{3}^{2}} \sqrt{12} = 3 \sqrt{12}$

$\sqrt{108} = \sqrt{{6}^{2} \cdot 3} = \sqrt{{6}^{2}} \sqrt{3} = 6 \sqrt{3}$

So there are several "simplifications" of $\sqrt{108}$.

When asked to simplify such a square root, what is normally expected is the expression that no longer has any square factors in the radicand. So in our example, $6 \sqrt{3}$ is considered simplified, whereas $2 \sqrt{27}$ or $3 \sqrt{12}$ are not fully simplified.

So if your teacher marked $2 \sqrt{27}$ (from $\sqrt{108} = \sqrt{4 \cdot 27}$) as being wrong, it was not because it was not equal (it is), but because it was not fully simplified.