# Square Root Property Formula

Mathematically, square is obtained when the number is multiplied by itself. But square root, is much more complicated to find the original number required. Which is why this formula is used. The required square number is usually a lengthy process and result in a long decimal form. We know that the square root 81 is 9, but what if we have to find the square root of 5? Using the calculator, we can see that the square root of 5 , which is denoted by: $\sqrt{5}$ equals 2.236. Square root sometimes need to found out in geometry to find the side of a particular shape. Let’s say, in pythagoras theorem, the result of finding the side using the theorem need not be in perfect square. However, we need to keep in mind the properties while finding out the square root.

Let’s see what those properties say:

$\sqrt{a}\cdot \sqrt{b}=\sqrt{a \times b}$

This says if we are multiplying the values individually, we might as well multiply them together for simplicity.

$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$

We can find the root of the denominator and numerator separately.

$\sqrt{n^{2}\cdot a}=n\sqrt{a}$

If a value is present in the root then that equals the square of the number, we should take it out of the root symbol and do our further calculation.

$\sqrt{a}+\sqrt{b}\neq \sqrt{a+b}$

If the values are in separate roots while addition, then they shouldn’t be clubbed together.

$\sqrt{a}-\sqrt{b}\neq \sqrt{a-b}$

If the values are in separate roots when subtracting, then they shouldn’t be clubbed together.

### Solved example

**Question:** Solve $\sqrt{81}+\sqrt{60}$ ?

We can’t club the terms together according to property number 4. Solving them individually we get,

$=\sqrt{9\times 9}+\sqrt{2\times 2\times 3\times 5}$

$=9+2\sqrt{15}$

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