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# nth Roots

You've previously learned that the square root of a number b can be written as $\sqrt{b}$ and satisfies the equation $b={x}^{2}$ . For example, $\sqrt{49}=7$ because ${7}^{2}=49$ . However, square roots are not the only roots you might encounter in mathematics.

Cubic roots are like square roots except that they are written as $\sqrt[3]{b}$ and satisfy the equation $b={x}^{3}$ . For instance, $\sqrt[3]{64}=4$ because ${4}^{3}=64$ . You can keep going up as well by adding a greater number on the outside of the $\sqrt{}$ symbol.

We can use mathematics to express this in generalized terms. The nth root of b can be written mathematically as $\sqrt[n]{b}$ and represents the number x where ${x}^{n}=b$ . Alternatively, you can write the nth root as a fractional exponent:

$\sqrt[n]{b}={b}^{\frac{1}{n}}$

## When do nth roots exist and how many are there?

Assuming you are working exclusively within the real number system:

If n is an even whole number, the nth root of b exists whenever b is positive and for all potential values of b.

If n is an odd whole number, the nth root of b exists for all potential values of b.

Therefore, $\sqrt[4]{-81}$ doesn't exist in the real number system because n is an even whole number (4) and b is not positive (-81). On the other hand, $\sqrt[3]{-8}$ exists because n is an odd whole number (3).

It's important to note that while the nth root of b may exist it won't always be a rational number. For example, consider $\sqrt{2}$ . Both n and b equal 2, and the number must exist since 2 is positive. However, $\sqrt{2}$ is an irrational number since it never terminates or falls into a repeating pattern in decimal form.

## When do nth roots exist in the complex number system and how many are there?

If you are working with the complex number system, things get much more complicated. Every number has two square roots, three cubic roots, four fourth roots, and so on without exception. Speaking in math terms, every number b has n nth roots. For instance, here are the four fourth roots of the number 81:

${3}^{4}=81$

${\left(-3\right)}^{4}=81$

${\left(3i\right)}^{4}={3}^{4}{i}^{4}=81$

${\left(-3i\right)}^{4}={\left(-3\right)}^{4}{i}^{4}=81$

So, the four fourth roots of 81 are $3,-3,3i$ and $-3i$ because they all satisfy the equation ${x}^{4}=81$ . Since half of them involve the imaginary unit, you may want to go back for a refresher on the topic if you're a little fuzzy.

## Practice questions on nth roots

a. What is $\sqrt{16}$ ?

4

b. What is $\sqrt[5]{-32}$ ?

-2

c. Does $\sqrt[6]{-27}$ exist in the real number system? Why or why not?

No, n is an even whole number and b is negative.

d. List all of the fourth roots of 256 in the complex number system.

$4,-4,4i,-4i$

## Flashcards covering the nth Roots

Algebra II Flashcards

## Varsity Tutors can help with nth roots

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