When simplifying square roots, begin by prime factoring the number in question. For \(\displaystyle \small 1260\), this is:

\(\displaystyle \small 1260=2*2*3*3*5*7\)

Now, for each pair of numbers, you can remove that number from the square root. Thus, you can say:

\(\displaystyle \small \sqrt{1260}=2*3*\sqrt{35}=6\sqrt{35}\)

Another way to think of this is to rewrite \(\displaystyle \small \sqrt{1260}\) as \(\displaystyle \small \sqrt{4}*\sqrt{9}*\sqrt{35}\). This can be simplified in the same manner.

We need to factor the number in the square root and find pairs of factors inorder to simplify a square root.

Since 83 is prime, it cannot be factored.

Thus the square root is already simplified.

If the triangle is a right triangle, then it follows the Pythagorean Theorem. Therefore:

\(\displaystyle a^2 + b^2 = c^2\) ---> \(\displaystyle 5^2 + 5^2 = 50 = c^2\)

\(\displaystyle c= \sqrt50\)

At this point, factor out the greatest perfect square from our radical:

\(\displaystyle \sqrt{50} = \sqrt{25 \cdot 2}\)

Simplify the perfect square, then repeat the process if necessary.

\(\displaystyle \sqrt{25 \cdot 2} = 5\sqrt2\)

Since \(\displaystyle 2\) is a prime number, we are finished!

There are two ways to solve this problem. If you happen to have it memorized that \(\displaystyle 2500\) is the perfect square of \(\displaystyle 50\), then \(\displaystyle \sqrt{5000} = \sqrt{2500 \cdot 2} = 50\sqrt 2\)   gives a fast solution.

If you haven't memorized perfect squares that high, a fairly fast method can still be achieved by following the rule that any integer that ends in \(\displaystyle 00\) is divisible by \(\displaystyle 25\), a perfect square.

\(\displaystyle \sqrt{5000} = \sqrt{25 \cdot 200} = 5\sqrt{200}\)

Now, we can use this rule again:

\(\displaystyle 5\sqrt{200} = 5\sqrt{25 \cdot 8} = 25 \sqrt {8}\)

Remember that we multiply numbers that are factored out of a radical.

The last step is fairly obvious, as there is only one choice:

\(\displaystyle 25 \sqrt{8} = 25\sqrt {4 \cdot 2} = 50\sqrt 2\)

A good method for simplifying square roots when you're not sure where to begin is to divide by \(\displaystyle 10\), \(\displaystyle 4\) or \(\displaystyle 16\), as one of these generally starts you on the right path. In this case, since our number ends in \(\displaystyle 0\), let's divide by \(\displaystyle 10\):

\(\displaystyle \sqrt{3240} = \sqrt{324 \cdot 10}\)

As it turns out, \(\displaystyle 324\) is a perfect square!

\(\displaystyle \sqrt{324 \cdot 10} = 18\sqrt{10}\)

Again here, if no perfect square is easily recognized try dividing by \(\displaystyle 10\), \(\displaystyle 4\), or \(\displaystyle 16\).

\(\displaystyle 8\sqrt{136} = 8\sqrt{4 \cdot 34} = 16 \sqrt{34}\)

Note that the \(\displaystyle 2\) we obtained by simplifying \(\displaystyle \sqrt{4}\) is multiplied, not added, to the \(\displaystyle 8\) already outside the radical.

To solve, simply find a perfect square factor and pull it out of the square root.

Recall the factors of 48 include (16, 3). Also recall that 16 is a perfect square since 4*4=16.

Thus,

\(\displaystyle \sqrt{48}\Rightarrow \sqrt{16*3}\Rightarrow \sqrt{4^2*3}\Rightarrow 4\sqrt{3}\)

The trick to these problems is to simplify the radical by using the following rule:  \(\displaystyle \sqrt{ab}=\sqrt{a}*\sqrt{b}\) and \(\displaystyle a\sqrt{b}+c\sqrt{b}=(a+c)\sqrt{b}\) Here, we need to find a common factor for the radical. This turns out to be five because \(\displaystyle 20=5*4\textup{ and }45=9*5.\) Remember, we want to include factors that are perfect squares, which are what nine and four are. Therefore, we can rewrite the equation as: \(\displaystyle \sqrt{20}+\sqrt{45}=2\sqrt{5}+3\sqrt{5}=5\sqrt{5}.\)

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

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.

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