√75 can be broken down to √25 * √3. Which simplifies to 5√3.

Rewrite what is under the radical in terms of perfect squares:

$\displaystyle x^{2}=x\cdot x$

$\displaystyle x^{4}=x^{2}\cdot x^{2}$

$\displaystyle x^{6}=x^{3}\cdot x^{3}$

Therefore, $\displaystyle \sqrt{a^{3}b^{4}c^{5}}= \sqrt{a^{2}a^{1}b^{4}c^{4}c^{1}}=ab^{2}c^{2}\sqrt{ac}$.

We know that 25 is a factor of 50. The square root of 25 is 5. That leaves $\displaystyle \sqrt{2}$ which can not be simplified further.

Multiply by the conjugate and the use the formula for the difference of two squares:

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

$\displaystyle =$ $\displaystyle \frac{x + \sqrt{3}}{3x + \sqrt{2}}\cdot \frac{3x - \sqrt{2}}{3x - \sqrt{2}}$

$\displaystyle =$ $\displaystyle \frac{3x^{2} -x \sqrt{2} + 3x\sqrt{3} - \sqrt{6}}{(3x)^{2} - (\sqrt{2})^{2}}$ 

$\displaystyle =$ $\displaystyle \frac{3x^{2} -x \sqrt{2} + 3x\sqrt{3} - \sqrt{6}}{9x^{2} - 2}$

First find all of the prime factors of $\displaystyle 468$

$\displaystyle 468=6\ast78=6\ast6\ast13=2\ast3\ast2\ast3\ast13$

So $\displaystyle \sqrt{468}=\sqrt{2\ast2\ast3\ast3\ast13}=2\ast3\sqrt{13}=6\sqrt{13}$

$\displaystyle \sqrt{432}$

1. We know that $\displaystyle 432=(144)(3)$, which we can separate under the square root:

$\displaystyle \sqrt{144\cdot 3}$

2. 144 can be taken out since it is a perfect square: $\displaystyle 12\cdot12=144$. This leaves us with:

$\displaystyle 12\sqrt{3}$

This cannot be simplified any further.

When simplifying square roots, begin by prime factoring the number in question. For $\displaystyle \small 420$, this is:

$\displaystyle \small 420 = 2*2*3*5*7$

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

$\displaystyle \small \small \sqrt{420}=2\sqrt{105}$

Another way to think of this is to rewrite $\displaystyle \small \small \sqrt{420}$ as $\displaystyle \small \sqrt{4}*\sqrt{105}$. This can be simplified in the same manner.

When simplifying square roots, begin by prime factoring the number in question. This is a bit harder for $\displaystyle \small 700700$. Start by dividing out $\displaystyle \small 100$:

$\displaystyle \small 700700 = 7007 * 100$

Now, $\displaystyle \small 7007$ is divisible by $\displaystyle 7$, so:

$\displaystyle \small \small 700700 =1001 * 7 * 2*2*5*5$

$\displaystyle \small 1001$ is a little bit harder, but it is also divisible by $\displaystyle \small 7$, so:

$\displaystyle \small 700700 =143*7 * 7 * 2*2*5*5$

With some careful testing, you will see that $\displaystyle \small 143 = 11 * 13$

Thus, we can say:

$\displaystyle \small \small 700700 = 2*2*5*5*7 * 7*11*13$

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

$\displaystyle \small \small \sqrt{700700} = 2*5*7*\sqrt{143}=70\sqrt{143}$

Another way to think of this is to rewrite $\displaystyle \small \small \small \sqrt{700700}$ as $\displaystyle \small \sqrt{4} * \sqrt{25} * \sqrt{49} * \sqrt{143}$. This can be simplified in the same manner.

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.