# Product, division and sum of square roots

## Product and division

The square root of the product of two numbers is the product of two square roots of the previously mentioned numbers, that is to say: $$\sqrt{x\cdot y}=\sqrt{x}\cdot\sqrt{y}$$$$$\sqrt{36}=\sqrt{4\cdot9}=\sqrt{4}\cdot\sqrt{9}=2\cdot3=6$$ or also $$\sqrt{25\cdot81}=\sqrt{25}\cdot\sqrt{81}=5\cdot9=45$$ The square root of a quotient is the quotient of the square roots, that is to say: $$\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{x}}{\sqrt{y}}$$$

$$\sqrt{\dfrac{16}{4}}=\dfrac{\sqrt{16}}{\sqrt{4}}=\dfrac{4}{2}=2$$

or also

$$\sqrt{\dfrac{49}{64}}=\dfrac{\sqrt{49}}{\sqrt{64}}=\dfrac{7}{8}$$

These two properties facilitate the calculation of roots of numbers that are the product of two perfect squares.

If we want to compute the root of $$11.664$$, knowing that $$144\cdot81=11.664$$ we can solve it easily.

$$\sqrt{11.664}=\sqrt{144\cdot81}=\sqrt{144}\cdot\sqrt{81}=12\cdot9=108$$

This also makes the calculation of roots of quotients easier. For instance,

$$\sqrt{\dfrac{784}{625}}=\dfrac{\sqrt{784}}{\sqrt{625}}=\dfrac{\sqrt{16\cdot49}}{\sqrt{25\cdot25}}=\dfrac{\sqrt{16}\sqrt{49}}{\sqrt{25}\sqrt{25}}=\dfrac{4\cdot7}{5\cdot5}=\dfrac{28}{25}$$

## Sum of square roots

We must be aware that the square root of the sum of two numbers is not the same as the sum of the roots of the original numbers, that is: $$\sqrt{9+4}\neq \sqrt{9}+\sqrt{4}$$$because if we compute it we have: $$\sqrt{9+4}=\sqrt{13}$$$ on one hand and $$\sqrt{9}+\sqrt{4}=3+2=5$$\$ on the other hand.

So we have established that: $$\sqrt{13}=5$$ and that is impossible since $$5\cdot5$$ is not $$13$$.

When we are managing an expression like: $$\sqrt{x}+\sqrt{y}$$ the roots CANNOT be put together and written as$$\sqrt{x}+\sqrt{y}=\sqrt{x+y}$$.

However, when we have an expression of the type $$a\sqrt{x}+b\sqrt{x}$$ it is allowed to add (and to subtract also) the coefficients.

$$5\sqrt{17}-2\sqrt{17}=3\sqrt{17}$$

Let's see another example:

We want to compute

$$5\dfrac{\sqrt{144}}{\sqrt{16}}+7\sqrt{21}-\sqrt{64}+2\sqrt{21}$$

First, using the most common perfect squares, we know that $$\sqrt{144}=12$$, $$\sqrt{16}=4$$, $$\sqrt{64}=8$$ and we replace it in the expression that we want to solve:

$$5\dfrac{12}{4}+7\sqrt{21}-8+2\sqrt{21}$$

Now we compute the sums between the numbers that have $$\sqrt{21}$$ on the one hand, and those that do not on the other.

$$\Big(5\dfrac{12}{4}-8\Big)+(7+2)\sqrt{21}$$

Computing, the result is:

$$7+9\sqrt{21}$$