Combining Powers and Roots

The square root is the opposite operation to exponentiation, and exponents are the opposite operation to square roots.
It's not for nothing that we will encounter a lot of exercises in a perfect combination, and we must know very well how to maneuver between the two.
That's exactly why we are here to teach you rules that will help you combine roots and powers.
Shall we begin?

Let's start with the first property and the basics.

Square root means a power of $0.5$.
Let's formulate it this way:
$\sqrt a=a^{ 1 \over 2}$
For example:
$√5=5^{0.5 }$

Each root has its own order. An order that appears in the root will translate into a denominator when the numerator has a share in the denominator of the number, if any.

$\sqrt[n]{a^m}=a^{\frac{m}{n}}$

$\sqrt[3]{9}=9^{\frac{1}{3}}$

Root of a Product
If we are given two numbers, which include a multiplication operation with a root of the same order, we can write a root that will cover the total product of the elements with the order that appears.
This rule can also help us to make a product of two factors with a root for two separate factors that have a root and a multiplication operation between them.

Let's formulate it this way:
$\sqrt{(a\times b)}=\sqrt{a}\times \sqrt{b}$

$√3\times √5=√15$

Let's translate this into powers:
$\sqrt{3}\times \sqrt{5}=3^{\frac{1}{2}}\times 5^{\frac{1}{2}}$
Similarly, we can say that:
$3^{\frac{1}{2}}\times 5^{\frac{1}{2}}=(3\times 5)^{\frac{1}{2}}=15^{\frac{1}{2}}=\sqrt{15}$

Root of a Quotient
If we are given two numbers, which include a division operation (fraction line) and a root with the same index, we can write a root that will be over each quotient of the elements with the index that appears.
This rule can also help us to make a quotient of two factors with a separate root into two factors that have a root and a division operation between them: a fraction line.

Let's put it this way:
$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$

$\frac{\sqrt[5]{7}}{\sqrt[5]{12}}=\sqrt[5]{\frac{7}{12}}$

Root of a Root
When we encounter an exercise where there is a root within a root, we can multiply the index of the first root by the index of the second root, and the index we obtain will be executed as a single root over our number. (As in the rule of power to a power)
Let's put it this way: 
$\sqrt[n]{\sqrt[m]{a}}=\sqrt[n\times m]{a}$

$\sqrt[5]{\sqrt[2]{20}}=\sqrt[10]{20}$

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If you're interested in this article, you might also be interested in the following articles:

Laws of Radicals

The Root of a Product

Root of a Quotient

Radication

On the Tutorela blog, you'll find a variety of articles about mathematics.

Assignment

What value should we place to solve the following equation?

$7^{\square}=49$

To answer this question, it is possible to respond in two ways:

One way is by substitution:

We place a power of $2$ and it seems that we have arrived at the correct result, that is:

$7²=49$

Another way is through the square root

$\sqrt{49}=7$

That is

$7²=49$

Answer:

$2$

Assignment

Which of the following clauses is equivalent to the expression:

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

Solution

According to the properties of square roots

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

Answer

Assignment

What is the answer to the exercise?

$\sqrt{3}\cdot\sqrt{12}+3^2$

Solution

According to the properties of square roots

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

$\sqrt{3}\cdot\sqrt{12}=\sqrt{3\cdot12}$

$\sqrt{36}=6$

Therefore

$\sqrt{3}\cdot\sqrt{12}+3^2=$

$6+3^2=6+9=15$

Answer

$15$

Assignment

Calculate and determine the answer:

$(\sqrt{9}-\sqrt{4})^2\cdot4^2-5^1$

Solution

We start with the parentheses

$\left(\sqrt{9}-\sqrt{4}\right)^2=\left(3-2\right)^2$

$1^2=1$

and therefore the following equation

$1^2\cdot4^2-5^1=$

$1\cdot16-5=$

$11$

Answer

$11$

Assignment

Calculate and determine the answer:

$(4^2+3^2):\sqrt{25}$

Solution

We start with the parentheses

$\left(4^2+3^2\right)=$

$16+9=25$

and then we calculate

$25:\sqrt{25}=$

$25:5=5$

Answer

$5$

Assignment

Calculate and determine the answer:

$(3^2+2^2)^2:(\sqrt{256}-\sqrt{9})-\sqrt{9}\cdot\sqrt{9}$

Solution

We start with the first parentheses

$(3^2+2^2)^2=$

$(9+4)^2=$

$13^2=169$

We calculate the second parentheses

$\left(\sqrt{256}-\sqrt{9}\right)=$

$16-3=13$

We calculate the expression after the subtraction

$\sqrt{9}\cdot\sqrt{9}=9$

and then we obtain

$169:13-9=13-9=4$

Answer

$4$

The elements of a root are $4$: the index of the radical, the radical sign, the radicand, and the root.

Index of the radical: It is the number that is outside and above the radical sign, indicating the number of times the root must be multiplied to obtain the number inside the radical sign.

Radical sign: The symbol for the radical operation \sqrt{\placeholder{}}

Radicand: It is the number inside the radical sign, which is the number from which the root will be extracted.

Root: It is the result of the radical operation.

With these elements, we can now define the root, and as we have said, it is the result. When we raise the result to the power indicated by the index, we will get the radicand, that is, the number inside the radical sign.

$\sqrt[3]{125}=5$

In this example, the radical index is $3$

The radicand is $125$

And the root is $5$

This means that if we raise $5$ to the power of $3$, we will get $125$, in other words;

$5^3=125$

As we know, all operations have an inverse operation. We know that the inverse operation of addition is subtraction and vice versa, for multiplication it is division, and for roots, their inverse operations are powers. This is how they are related as inverse operations. Let's see this relationship with some examples:

$\sqrt[3]{64}$

That is, we must find a number that, when multiplied by itself $3$ times, gives us $64$. From this, we can deduce that the root will be $4$, since we know the following:

$4^3=64$

Therefore, the result is

$4$

Calculate the following

$\sqrt{144}=$

Here we do not have the index of the radical explicitly shown, but when this happens and no index is visible, we assume that the index is a $2$. So we need to find a number that, when multiplied by itself twice, gives us $144$. In this case, the answer is $12$, since:

$12^2=144$

Therefore, the result is

$12$

To solve combined calculations with roots and powers, we must first take into account the order of operations and then the laws and properties of powers and roots.

Solve

\(\left(\sqrt[3]{8}+\sqrt{100}\right)^2-4^2+\sqrt{81}=

By the order of operations, we solve the grouping sign, which are the parentheses, and we can do it separately, as follows:

$\left(\sqrt[3]{8}+\sqrt{100}\right)^2=\left(2+10\right)^2=\left(12\right)^2=144$,$4^2=16$

$\sqrt{81}=9$

According to this then we have the following:

$144-16+9=137$

Therefore, the result is

$137$

We must remember that there is a hierarchy of operations (order in which operations should be performed). The order is as follows:

1. Grouping symbols (parentheses, brackets, and braces)
2. Powers and roots
3. Multiplication and division
4. Addition and subtraction

When we encounter operations that have the same rank, such as powers and roots, and when they appear in combination, they are solved from left to right

$\sqrt{16}+8-\left(3\right)^2=4+8-9=3$

In this example, we can see that we have a square root, an addition, and a subtraction of a power. Since the square root and the power are independent, they can be performed at the same time, and finally, we carry out the addition and subtraction.

$\sqrt{19+\left(9\right)^2}=\sqrt{19+81}=\sqrt{100}=10$

Here we can see that there is a square root and an exponent, so we first solve the square root but inside the square root we have an exponent, therefore we must first solve the exponent $9^2=81$, then we proceed with the addition and finally we calculate the square root.

There are $5$ types of radical rules, which are called the laws of radicals, and they are as follows:

• First law

$\sqrt{a}=a^{\frac{1}{2}}$

• Second law

$\sqrt[n]{a^m}=a^{\frac{m}{n}}$

• Third law

$\sqrt[n]{a\times b}=\sqrt[n]{a}\times\sqrt[n]{b}$

• Fourth law

$\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$

• Fifth law

$\sqrt[n]{\sqrt[m]{a}}=\sqrt[n\times m]{a}$