\(\displaystyle \sqrt[2]{24,300}\)

When simplifying the square root of a number that may not have a whole number root, it's helpful to approach the problem by finding common factors of the number inside the radicand. In this case, the number is 24,300.

What are the factors of 24,300?

24,300 can be factored into:

\(\displaystyle 24,300: 243\cdot 100\)

When there are factors that appear twice, they may be pulled out of the radicand. For instance, 100 is a multiple of 24,300. When 100 is further factored, it is \(\displaystyle 10^{2}\) (or 10x10). However, 100 wouldn't be pulled out of the radicand, but the square root of 100 because the square root of 24,300 is being taken. The 100 is part of the24,300. This means that the problem would be rewritten as: \(\displaystyle 10\sqrt[2]{243}\) 

But 243 can also be factored: \(\displaystyle 24,300: (3\cdot 81)\cdot100\)
\(\displaystyle 24,300: {\color{Orange} 9}\cdot{\color{Orange} 9}\cdot {\color{Cyan} 10}\cdot {\color{cyan} 10}\cdot 3\)
Following the same principle as for the 100, the problem would become
\(\displaystyle 9\cdot 10\sqrt[2]{3}\) because there is only one factor of 3 left in the radicand. If there were another, the radicand would be lost and it would be 9*10*3. 
9 and 10 may be multiplied together, yielding the final simplified answer of 
\(\displaystyle 90\sqrt[2]{3}\)

To solve the equation \(\displaystyle \sqrt{180} + \sqrt{125} = ?\), we can first factor the numbers under the square roots.

\(\displaystyle \sqrt{2\cdot 2\cdot 3\cdot 3\cdot 5} + \sqrt{5\cdot 5\cdot 5} = ?\)

When a factor appears twice, we can take it out of the square root.

\(\displaystyle 2\cdot 3\sqrt{5} + 5\sqrt{5} = ?\)

\(\displaystyle 6\sqrt{5} + 5\sqrt{5} = ?\)

Now the numbers can be added directly because the expressions under the square roots match.

\(\displaystyle 6\sqrt{5} + 5\sqrt{5} = 11\sqrt{5}\)

To simplify, we must try to find factors which are perfect squares. In this case 16 is a factor of 624 and is also a perfect square.

Therefore we can rewrite the square root of 624 as:

\(\displaystyle \sqrt{624}=\sqrt{16\times 39}=\sqrt{16}\sqrt{39}=4\sqrt{39}\)

To simplify, we must try to find factors which are perfect squares. In this case 20 is a factor of 400 and is also a perfect square.

Thus we can rewrite the problem as:

\(\displaystyle \sqrt{400}=\sqrt{20\times 20}=20\)

Note: \(\displaystyle 20^{2}=20\times20=400\)

Use the following steps to reduce this square root.

To simplify, we must try to find factors which are perfect squares. In this case 144 is a factor of 720 and is also a perfect square.

Thus we can rewrite the problem as follows.

\(\displaystyle \sqrt{720}=\sqrt{144\times 5}=\sqrt{144}\sqrt{5}=12\sqrt{5}\)

Use the following steps to find the square root of \(\displaystyle 1,800:\)

To simplify, we must try to find factors which are perfect squares. In this case 900 is a factor of 1800 and is also a perfect square.

Thus we can rewrite the problem as follows.

\(\displaystyle \sqrt{1,800}=\sqrt{900\times 2}=\sqrt{900}\sqrt{2}=30\sqrt{2}\)

To simplify, we must try to find factors which are perfect squares. In this case 9 is a factor of 54 and is also a perfect square.

To reduce this expression, use the following steps: 

\(\displaystyle \sqrt{54}=\sqrt{9\times 6}=\sqrt{9}\sqrt{6}=3\sqrt{6}\)

To simplify, we must try to find factors which are perfect squares. In this case 36 is a factor of 72 and is also a perfect square.

To reduce this expression, use the following arithmetic steps: 

\(\displaystyle \sqrt{72}=\sqrt{36\times 2}=\sqrt{36}\sqrt{2}=6\sqrt{2}\)

To simplify, we must try to find factors which are perfect squares. In this case 30 is a factor of 900 and is also a perfect square.

The square root of \(\displaystyle 900\) is equal to: 

\(\displaystyle \sqrt{900}=\sqrt{30\times 30}=30\)

However,

\(\displaystyle \sqrt{900}=\sqrt{-30\times -30}=-30\)

Thus, \(\displaystyle \pm30\) \(\displaystyle =\sqrt{900}\)

To simplify, we must try to find factors which are perfect squares. In this case 16 is a factor of 32 and is also a perfect square.

To reduce this expression, use the following steps:

\(\displaystyle \sqrt{32}=\sqrt{16\times 2}=\sqrt{16}\sqrt{2}=4\sqrt{2}\)