# Simplify $\frac{\sqrt{24}}{8}$

The question I have been given is to simplify as much as possible: $$\frac{\sqrt {24}}{8}$$. I know the answer is $$\frac{\sqrt 6}{4}$$

(Note I am in a beginner math course, so go easy on me.)

My first thought was to divide to get: $$\frac{\sqrt {12}}{4}$$ and then again: $$\frac{\sqrt 3}{1}$$. However, I realized this was wrong. So I tried $$\frac{\sqrt {24}}{2\times\sqrt 4}$$ which would make the denominator equal to 4, which is right. So I thought I could do that to the top too, but I couldn't. I feel like I'm on the right track but not really there. Can someone help me figure out how to solve questions like these?

• Dividing by 2 is ok, but then the numerator would read: $$\sqrt{24}/2 = \sqrt{24}/\sqrt{2^2} =\sqrt{24/4}.$$ Since the rule is $\sqrt{ab} = \sqrt{a}\sqrt{b}$ Sep 11, 2017 at 8:12
• You cannot cancel $\frac{\sqrt{24}}{8}$ the same way as you do for $\frac{24}{8}$. One has to respect the square root action. Note that $\frac{\sqrt{2}}{2}\ne \frac11=1$.
– A.Γ.
Sep 11, 2017 at 8:14

You could square the fraction, then simplify, then take the square root. $$\frac{\sqrt{24}^2}{8^2}=\frac{24}{64}=\frac{6}{16}$$

Taking the square root of both sides leaves the desired answer. $$\frac{\sqrt{6}}{\sqrt{16}}=\frac{\sqrt{6}}{4}$$

You can see $\sqrt{24} = \sqrt{4 \cdot 6}$. Since $\sqrt{a\cdot b} = \sqrt{a} \cdot \sqrt{b}$, you can expand the nominator to result in $2\cdot \sqrt{6}$. You can then divide by the $2$ to get the desired result.

• Thanks! I understand now :D Sep 11, 2017 at 8:13

You should first simplify the square root part as much as possible, then cancel the fraction.

$$\sqrt{24}=\sqrt{4\times 6}=\sqrt{4}\times\sqrt{6}=2\sqrt{6}$$

Here you should always be trying to make one of the factors the largest square number that divides the number you are taking the square root of (here $4$ is the largest square that divides $24$).

Now you can simplify the fraction: $$\frac{\sqrt{24}}{8}=\frac{2\times\sqrt{6}}{8}=\frac{1\times\sqrt{6}}4=\frac{\sqrt{6}}4.$$

when doing this you can only cancel numbers from the top and bottom if they are both outside the square root (or both inside).

• Great explanation, thank you! :) Sep 11, 2017 at 8:14

How would you go about simplifying $$\frac{24}{8}?$$ You could try repeatedly halving the numerator and the denominator until arriving at the answer: $$\frac{24}{8} = \frac{12}{4} = \frac{6}{2} = \frac{3}{1} = 3.$$

Believe it or not, $$\frac{\sqrt{24}}{2} = \sqrt 6.$$ But hey, I'm a demon, I could be trying to lead you astray. So check it on your calculator.

So, by the same process as the other one, $$\frac{\sqrt{24}}{8} = \frac{\sqrt{6}}{4}.$$

The radicand, $$24$$, can be expressed as a product involving a perfect square and a non-perfect square. In this case $$24$$ is equal to $$4$$ (which is a perfect square) times $$6$$ (a non-perfect square). Thus

$$\sqrt{24} = \sqrt{4 \times 6} = \sqrt{4}\sqrt{6} = 2\sqrt{6}$$ and $$\frac{\sqrt{24}}{8} = \frac{2\sqrt{6}}{8}$$

After dividing both the numerator and denominator by the common factor of $$2$$ and $$8$$, which is $$2$$:

$$\frac{2\sqrt{6} / 2}{8/2} = \frac{\sqrt{6}}{4}$$