To simplify a square root, you have to factor the number and look for pairs. Whenever there is a pair of factors (for example two twos), you pull one to the outside.

Thus when you factor 96 you get
$\displaystyle \\ \sqrt{96}=\sqrt{2\cdot 2\cdot 2\cdot2\cdot2\cdot 3}\\ \\=2\cdot 2\sqrt{2\cdot 3}\\ \\=4\sqrt{6}$

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: $\displaystyle \sqrt{20}+\sqrt{45}=2\sqrt{5}+3\sqrt{5}=5\sqrt{5}.$

This question may look daunting, especially if you start out by trying pick values or $\displaystyle x$ and solving. This plan of attack will work, but you're likely going to be dealing with some messy numbers. Instead we want to recall some of our exponent and root rules. 

Let's look at this answer choice: 

$\displaystyle \sqrt{(\sqrt{x^3})}$

This double square root is the same as a fourth root. Think about it, you have a square times a square- which is the same thing as $\displaystyle 2\times2$. Thus,  $\displaystyle \sqrt{(\sqrt{x^3})}=\sqrt[4]{x^3}$ so this choice can be eliminated. 

Next, let's look at this answer choice: 

$\displaystyle x^{\frac{3}{4}}$

For this choice, we need to recall our exponent rules. Remember, whenever we have a value raised to a fractional power, the denominator of that fraction is equal to the root number. In this case, $\displaystyle 4$. Thus, $\displaystyle x^{\frac{3}{4}}=\sqrt[4]{x^3}$

Now, let's look at this answer choice:

$\displaystyle (\sqrt[4]{x})^3$

A key rule to remember here is that order doesn't not matter when dealing with roots and powers. Thus, taking the $\displaystyle 4^{th}}$ root of a number and then cubing it will result in the same value as cubing a number and then taking the $\displaystyle 4^{th}}$ root . 

This leaves us with: 

$\displaystyle \sqrt{x^{\frac{2}{3}}}$

If we tried to break this down a bit, we could take the third root of $\displaystyle x^2$, which would leave us with: 

$\displaystyle \sqrt{\sqrt[3]{x^2}}$. This will not equal $\displaystyle \sqrt[4]{x^3}$.