Square Root Rules

A root is symbolized with the sign $√$
Indeed, when we see a number with a root, we wonder what positive number raised to $2$, will give us what is written inside the root.
A root is the opposite of a power operation. When there is no small number at the top left of the root, it denotes that it is a root of $2$, square root.
If a small number appears on the left, this will be the order of the root.

Let's know some of the definitions and properties:

1. The result of a root will always be positive! A negative result will never be obtained. We can get a result of $0$.
2. For $√$ (negative number) there is no answer!
3. The root is basically a half power. We can say that: $\sqrt a=a^{ 1 \over 2}$
4. In the order of Operations, square roots are evaluated along with other exponents in the standard order of operations (PEMDAS/BODMAS), after parentheses and before multiplication and division.

Let's see the example:
$\sqrt {64} =8$

Let's ask, w What number times itself will give us $64$ and the answer is $8$.
True, also $-8$ to the power of $2$ will give us $64$ but the result of the root must be positive!

The following laws allow you to manipulate and simplify radical expressions. These rules are based on the properties of exponents and are essential for solving complex radical problems.

When the root appears across the entire product, we can break down each factor and apply the root to them, leaving the multiplication sign between the factors.
We formulate:
$\sqrt{(a\cdot b)}=\sqrt{a}\cdot\sqrt{b}$[object Object]
Where $a \geq 0$ and $b \geq 0$

Let's see this in an example:
$\sqrt{(64\cdot100)}$
According to the root of a product rule, we can break down the factors and leave the root of each factor separately while maintaining the multiplication operation between them:
We will break it down and obtain:
$\sqrt{64}\cdot\sqrt{100}=$
$8\cdot10=80$

More Examples:

• $\sqrt{4 \cdot 9} = \sqrt{4} \cdot \sqrt{9} = 2 \cdot 3 = 6$
• $\sqrt{50} = \sqrt{25 \cdot 2} = \sqrt{25} \cdot \sqrt{2} = 5\sqrt{2}$
• $\sqrt{3} \cdot \sqrt{12} = \sqrt{3 \cdot 12} = \sqrt{36} = 6$

Key Application: Use this rule to simplify radicals by factoring out perfect squares.

When the root appears over the entire quotient (over the entire fraction), we can break down each factor and apply the root to it, leaving the division sign (fraction line) between the factors.
We formulate:
$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$[object Object]
where $a \geq 0$ and $b > 0$

Let's see this in an example:

$\sqrt{\frac{36}{9}}$

According to the rule of the root of a quotient, we can break down the factors and leave the root of each factor separately while maintaining the multiplication operation between them:
We will break it down and obtain:

$\frac{\sqrt{36}}{\sqrt{9}}$

$\frac{6}{3}=2$

More Examples:

• $\sqrt{\frac{25}{4}} = \frac{\sqrt{25}}{\sqrt{4}} = \frac{5}{2}$
• $\sqrt{\frac{18}{2}} = \frac{\sqrt{18}}{\sqrt{2}} = \frac{3\sqrt{2}}{\sqrt{2}} = 3$
• $\frac{\sqrt{50}}{\sqrt{2}} = \sqrt{\frac{50}{2}} = \sqrt{25} = 5$

Important Note: This rule is commonly used for rationalizing denominators and simplifying complex fractions.

When we encounter an exercise where there is a root over a root, we will multiply the order of the first root by the order of the second root and the order we obtain we will multiply as a root over our number. (As the rule of power over another power)
Let's put it this way:
$\sqrt[n]{\sqrt[m]{a}}=\sqrt[n\cdot m]{a}$[object Object]
Where $a \geq 0$.

Let's see this in the example: 
$\sqrt[2]{\sqrt[4]{100}}=\sqrt[2\cdot4]{100}=\sqrt[8]{100}$

More Examples:

• $\sqrt{\sqrt{16}} = \sqrt[4]{16} = 2$ (since $2^4 = 16$)
• $\sqrt{\sqrt[3]{8}} = \sqrt[6]{8} = 8^{1/6}$
• $\sqrt{\sqrt{x^8}} = \sqrt[4]{x^8} = x^2$ (when $x \geq 0$)

When working with multiple radicals, apply the rules systematically:

Step-by-Step Process:

1. Look for perfect square factors
2. Apply product/quotient rules
3. Combine like radicals
4. Rationalize denominators if needed

Examples:

Example 1: $\sqrt{72} + \sqrt{18}$

• $\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}$
• $\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}$
• Result:$6\sqrt{2} + 3\sqrt{2} = 9\sqrt{2}$

Example 2: $\frac{\sqrt{45}}{\sqrt{5}}$

• Method 1: $\frac{\sqrt{45}}{\sqrt{5}} = \sqrt{\frac{45}{5}} = \sqrt{9} = 3$
• Method 2: $\sqrt{45} = \sqrt{9 \cdot 5} = 3\sqrt{5}$, so$\frac{3\sqrt{5}}{\sqrt{5}} = 3$

Mistake 1: $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$

• Wrong: $\sqrt{9 + 16} = \sqrt{9} + \sqrt{16} = 3 + 4 = 7$
• Correct: $\sqrt{9 + 16} = \sqrt{25} = 5$

Mistake 2: Forgetting absolute values with variables

• Wrong:$\sqrt{x^2} = x$
• Correct: $\sqrt{x^2} = |x|$

Mistake 3: Incorrect application of quotient rule

• Wrong: $\sqrt{\frac{a}{b}} = \frac{a}{\sqrt{b}}$
• Correct: $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$

Mistake 4: Forgetting domain restrictions

• Remember: $\sqrt{a}$ is only defined for $a \geq 0$ in real numbers

If you are interested in this article, you may also be interested in the following articles:

The root of a product

Root of the quotient

Radication

Combining powers and roots

In the blog of Tutorela you will find a variety of articles on mathematics.