Reference

Properties of Radicals

Rule

$n^{th}$ Roots of $n^{th}$ Powers

To simplify an $n^{th}$ root, the radicand must first be expressed as a power. If the index of the radical and the power of the radicand are equal such that $n a^{n},$ the radical expression can be simplified as follows.

n a^{n} = {∣ a, ∣ a ∣, if n is odd if n is even

The absolute value of a number is always non-negative, so when $n$ is even, the result will always be non-negative. Consider a few example $n^{th}$ roots that can be simplified by using the formula.

$a$	$n$	$n a^{n}$	Is $n$ even?	Simplify
$4$	$3$	$3 4^{3}$	$\times$	$4$
$- 6$	$5$	$5 (- 6)^{5}$	$\times$	$- 6$
$2$	$6$	$6 2^{6}$	$✓$	$∣ 2 ∣ = 2$
$- 3$	$4$	$4 (- 3)^{4}$	$✓$	$∣ - 3 ∣ = 3$

Proof

Informal Justification

To write the expression for $n a^{n},$ there are two cases to consider.

$a \geq 0$
$a < 0$

Both cases will be considered one at a time.

$a \geq 0$

Start by noting that since

a

is non-negative,

a^{n}

is also non-negative. This means that the

n^{th}

root can be rewritten using a rational exponent.

n a^{n} = (a^{n})^{\frac{1}{n}}

Since both

n

and

\frac{1}{n}

are rational numbers, the Power of a Power Property for Rational Exponents can be applied to simplify the obtained expression.

(a^{n})^{\frac{1}{n}}

▼

Simplify

PowPow

$(a^{m})^{n} = a^{m \cdot n}$

a^{n \cdot \frac{1}{n}}

$a \cdot \frac{1}{a} = 1$

a^{1}

ExponentOne

$a^{1} = a$

a

Therefore,

n a^{n}

is equal to

a

a

is non-negative.

$a < 0$

In case of a negative value of $a,$ there are also two cases two consider.

$n$ is even
$n$ is odd

Even $n$

Recall that a root with an even index

n_{e}

is defined only for non-negative numbers. Although

a

is negative,

a^{n_{e}}

is positive. Also, a power with a negative base and an even exponent can be rewritten as a power with a positive base.

a^{n_{e}} = (- a)^{n_{e}}

Now, since

- a

is positive, the Power of a Power Property for Rational Exponents can be applied again to simplify

n_{e} a^{n_{e}} .

n_{e} a^{n_{e}}

▼

Simplify

NegBaseToPosPow

$(- a)^{n_{e}} = a^{n_{e}}$

n_{e} (- a)^{n_{e}}

RootToPowD

$n a = a^{\frac{1}{n}}$

((- a)^{n_{e}})^{\frac{1}{n _{e}}}

PowPow

$(a^{m})^{n} = a^{m \cdot n}$

(- a)^{n_{e} \cdot \frac{1}{n _{e}}}

$a \cdot \frac{1}{a} = 1$

(- a)^{1}

ExponentOne

$a^{1} = a$

- a

Odd $n$

A root with an odd index

n_{o}

is defined for all real numbers. By the definition of the

n^{th}

root, the expression

n_{o} a^{n_{o}}

is the number

y

that, when multiplied by itself

n_{o}

times, will result in

a^{n_{o}} .

n_{o} times y \cdot y \cdot \dots \cdot y = a^{n_{o}}

Because

n_{o}

is odd and

a

is negative,

a^{n_{o}}

is also negative. This means that the best candidate for

y

is simply

a .

n a^{n} = a

Summary

If $a$ is non-negative, $n a^{n}$ is always equal to $a .$ However, in case of negative $a,$ the value of $n a^{n}$ depends on the parity of $n .$

	$a \geq 0$	$a < 0$
Even $n$	$n a^{n} = a$	$n a^{n} = - a$
Odd $n$	$n a^{n} = a$	$n a^{n} = a$

To conclude, for odd values of $n,$ the expression $n a^{n}$ is equal to $a .$ On the other hand, if $n$ is even, $n a^{n}$ can be written as $∣ a ∣ .$

n a^{n} = {∣ a, ∣ a ∣, if n is odd if n is even

Extra

Simplifying

n a^{m}

Depending on the index of the root and the power in the radicand, simplifying $n a^{m}$ may be problematic. Because real $n^{th}$ roots with an even index are defined only for non-negative numbers, the absolute value is sometimes needed.

If $n$ is even, $n a^{m}$ is defined only for non-negative $a^{m} .$

The following applet presents a decision tree to simplify

n a^{m} .

In this applet,

k

is the greatest common factor of

m

and

n .

Consider example

n^{th}

roots

n a^{m}

that can be simplified by using the decision tree.

$a$	$m$	$n$	$n a^{m}$	Simplify
$2$	$8$	$4$	$4 2^{8}$	$2^{\frac{8}{4}} = 2^{2} = 4$
$- 3$	$6$	$2$	$(- 3)^{6}$	$∣ ∣ ∣ ∣ (- 3)^{\frac{6}{2}} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ (- 3)^{3} ∣ ∣ ∣ = ∣ - 27 ∣ = 27$
$x$	$6$	$3$	$3 x^{6}$	$x^{\frac{6}{3}} = x^{2}$
$- 3$	$2$	$8$	$8 (- 3)^{2}$	$4 ∣ - 3 ∣ = 43$
$2$	$3$	$6$	$6 2^{3}$	$2$
$- 2$	$3$	$9$	$9 (- 2)^{3}$	$3 - 2$
$2$	$6$	$8$	$8 2^{6}$	$4 2^{3} = 48$
$- 2$	$6$	$8$	$8 (- 2)^{6}$	$4 ∣ - 2 ∣^{3} = 4 2^{3} = 48$

Rule

Product Property of Radicals

Given two non-negative numbers $a$ and $b,$ the $n^{th}$ root of their product equals the product of the $n^{th}$ root of each number.

$n a b = n a \cdot n b,$ for $a \geq 0$ and $b \geq 0$

If $n$ is an odd number, the $n^{th}$ root of a negative number is defined. In this case, the Product Property of Radicals for negative $a$ and $b$ is also true.

Proof

First, a special case of the property will be proven. Assume that

a

b

is equal to

0 .

By the Zero Property of Multiplication, the radicand in left-hand side of the formula is equal to

0 .

\underline{Left - Hand Side} n 0 = ? \underline{Right - Hand Side} n a \cdot n b

Because

0^{n} = 0,

n^{th}

root of

0

is equal to

0 .

This means that both the left-hand side and one of the factors on the right-hand side equal

0 .

Again, by the Zero Property of Multiplication, the right-hand side is also

0 .

\underline{Left - Hand Side} 0 = \underline{Right - Hand Side} 0

Therefore, the property is true when

a

b

is equal to

0 .

Next, assume that both

a

and

b

are nonzero. Let

x,

y,

and

z

be real numbers such that

x = n a,

y = n b,

and

z = n a b .

By the definition of an

n^{th}

root, each of these numbers raised to the

n^{th}

power is equal to its corresponding radicand.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x^{n} = a y^{n} = b z^{n} = a b (I) (II) (III)

Next, because

b

is not equal to

0,

Equation (I) can be multiplied by

y^{n},

which is equal to

b .

x^{n} = a ⇓ x^{n} y^{n} = a y^{n}

Now, substitute Equations (II) and (III) into the obtained equation.

x^{n} y^{n} = a y^{n}

▼

Substitute values and simplify

Substitute

$y^{n} = b$

x^{n} y^{n} = a b

Substitute

$a b = z^{n}$

x^{n} y^{n} = z^{n}

ProdPowII

$a^{m} b^{m} = (a b)^{m}$

(x y)^{n} = z^{n}

RearrangeEqn

Rearrange equation

z^{n} = (x y)^{n}

Since

x y

and

z

are of the same sign, the final equation implies that

z = x y .

z^{n} = (x y)^{n} \Rightarrow z = x y

The last step is substituting

z = n a b,

x = n a,

and

y = n b

into this equation.

z = x y \Leftrightarrow n a b = n a \cdot n b

Rule

Quotient Property of Radicals

Let $a$ be a non-negative number and $b$ be a positive number. The $n^{th}$ root of the quotient $\frac{a}{b}$ equals the quotient of the $n^{th}$ roots of $a$ and $b .$

$n \frac{a}{b} = \frac{n a}{n b},$ for $a \geq 0,$ $b > 0$

If $n$ is an odd number, the $n^{th}$ root of a negative number is defined. In this case, the Quotient Property of Radicals for negative $a$ and $b$ is also true.

Proof

First, a special case of the property will be proven. Assume that

a

is equal to

0 .

Because

\frac{0}{b} = 0,

the radicand in left-hand side of the formula is equal to

0 .

\underline{Left - Hand Side} n 0 = ? \underline{Right - Hand Side} \frac{n 0}{n b}

Because

0^{n} = 0,

n^{th}

root of

0

is equal to

0 .

This means that both the left-hand side and the numerator on the right-hand side equal

0 .

Again, since dividing

0

by a nonzero number results in

0,

the right-hand side is also

0 .

\underline{Left - Hand Side} 0 = ✓ \underline{Right - Hand Side} 0

Therefore, the property is true when

a

is equal to

0 .

Next, assume that both

a

and

b

are nonzero. Let

x,

y,

and

z

be real numbers such that

x = n a,

y = n b,

and

z = n \frac{a}{b} .

By the definition of an

n^{th}

root, each of these numbers raised to the

n^{th}

power is equal to its corresponding radicand.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x^{n} = a y^{n} = b z^{n} = \frac{a}{b} (I) (II) (III)

Next, because

b

is not equal to

0,

Equation (I) can be divided by

y^{n},

which is equal to

b .

x^{n} = a ⇓ \frac{x ^{n}}{y ^{n}} = \frac{a}{y ^{n}}

Now, substitute Equations (II) and (III) into the obtained equation.

\frac{x ^{n}}{y ^{n}} = \frac{a}{y ^{n}}

▼

Substitute values and simplify

Substitute

$y^{n} = b$

\frac{x ^{n}}{y ^{n}} = \frac{a}{b}

Substitute

$\frac{a}{b} = z^{n}$

\frac{x ^{n}}{y ^{n}} = z^{n}

$\frac{a ^{m}}{b ^{m}} = (\frac{a}{b})^{m}$

(\frac{x}{y})^{n} = z^{n}

RearrangeEqn

Rearrange equation

z^{n} = (\frac{x}{y})^{n}

Since

\frac{x}{y}

and

z

are of the same sign, the final equation implies that

z = \frac{x}{y} .

z^{n} = (\frac{x}{y})^{n} \Rightarrow z = \frac{x}{y}

The last step is substituting

z = n \frac{a}{b},

x = n a,

and

y = n b

into this equation.

z = \frac{x}{y} \Leftrightarrow n \frac{a}{b} = \frac{n a}{n b}

Rule

Quotient Property of Radicals

Let $a$ be a non-negative number and $b$ be a positive number. The $n^{th}$ root of the quotient $\frac{a}{b}$ equals the quotient of the $n^{th}$ roots of $a$ and $b .$

$n \frac{a}{b} = \frac{n a}{n b},$ for $a \geq 0,$ $b > 0$

If $n$ is an odd number, the $n^{th}$ root of a negative number is defined. In this case, the Quotient Property of Radicals for negative $a$ and $b$ is also true.

Proof

First, a special case of the property will be proven. Assume that

a

is equal to

0 .

Because

\frac{0}{b} = 0,

the radicand in left-hand side of the formula is equal to

0 .

\underline{Left - Hand Side} n 0 = ? \underline{Right - Hand Side} \frac{n 0}{n b}

Because

0^{n} = 0,

n^{th}

root of

0

is equal to

0 .

This means that both the left-hand side and the numerator on the right-hand side equal

0 .

Again, since dividing

0

by a nonzero number results in

0,

the right-hand side is also

0 .

\underline{Left - Hand Side} 0 = ✓ \underline{Right - Hand Side} 0

Therefore, the property is true when

a

is equal to

0 .

Next, assume that both

a

and

b

are nonzero. Let

x,

y,

and

z

be real numbers such that

x = n a,

y = n b,

and

z = n \frac{a}{b} .

By the definition of an

n^{th}

root, each of these numbers raised to the

n^{th}

power is equal to its corresponding radicand.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x^{n} = a y^{n} = b z^{n} = \frac{a}{b} (I) (II) (III)

Next, because

b

is not equal to

0,

Equation (I) can be divided by

y^{n},

which is equal to

b .

x^{n} = a ⇓ \frac{x ^{n}}{y ^{n}} = \frac{a}{y ^{n}}

Now, substitute Equations (II) and (III) into the obtained equation.

\frac{x ^{n}}{y ^{n}} = \frac{a}{y ^{n}}

▼

Substitute values and simplify

Substitute

$y^{n} = b$

\frac{x ^{n}}{y ^{n}} = \frac{a}{b}

Substitute

$\frac{a}{b} = z^{n}$

\frac{x ^{n}}{y ^{n}} = z^{n}

$\frac{a ^{m}}{b ^{m}} = (\frac{a}{b})^{m}$

(\frac{x}{y})^{n} = z^{n}

RearrangeEqn

Rearrange equation

z^{n} = (\frac{x}{y})^{n}

Since

\frac{x}{y}

and

z

are of the same sign, the final equation implies that

z = \frac{x}{y} .

z^{n} = (\frac{x}{y})^{n} \Rightarrow z = \frac{x}{y}

The last step is substituting

z = n \frac{a}{b},

x = n a,

and

y = n b

into this equation.

z = \frac{x}{y} \Leftrightarrow n \frac{a}{b} = \frac{n a}{n b}

Method

Simplifying Radical Expressions

Expressions involving radicals be simplified using various properties of exponents and radicals. Consider the following square root.

50 x^{6} y^{4}

Follow these four steps to simplify the expression.

Factor the Radicand

Factor the number in the radicand into its prime factors and the variables into their smallest exponents. In this example, the number under the radical is

50

and it is a product of powers of

2

and

5 .

50 = 2 \cdot 5^{2}

Since

x^{6}

and

y^{4}

are already prime factors, the radical expression can be rewritten as follows.

50 x^{6} y^{4} = 2 \cdot 5^{2} \cdot x^{6} \cdot y^{4}

Separate Perfect Powers

Identify and separate perfect powers that match the index of the radical. Identifying perfect squares for a square root means recognizing expressions that can be rewritten as the product of two identical powers with integer exponents. In the given example,

5^{2},

x^{6},

and

y^{4}

are all perfect squares.

2 \cdot 5^{2} \cdot x^{6} \cdot y^{4} = 2 \cdot 5^{2} \cdot (x^{3})^{2} \cdot (y^{2})^{2}

Simplify the Perfect Powers

The product property of radicals states that for any non-negative numbers the root of their product equals the product of their roots. This allows radicals to be split into separate radicals. Next, simplify each perfect square radical by taking their square roots individually.

2 \cdot 5^{2} \cdot (x^{3})^{2} \cdot (y^{2})^{2}

SqrtProd

$a \cdot b = a \cdot b$

2 \cdot 5^{2} \cdot (x^{3})^{2} \cdot (y^{2})^{2}

SqrtPowToNumber

$a^{2} = a$

2 \cdot 5 \cdot (x^{3})^{2} \cdot (y^{2})^{2}

SqrtToAbs

$a^{2} = ∣ a ∣$

2 \cdot 5 \cdot ∣ x^{3} ∣ \cdot ∣ y^{2} ∣

Because

y^{2}

is always non-negative, the absolute value of

y^{2}

equals

y^{2} .

However, this rule does not hold for

x^{3}

because its exponent is odd.

2 \cdot 5 \cdot ∣ x^{3} ∣ \cdot ∣ y^{2} ∣ = 2 \cdot 5 \cdot ∣ x^{3} ∣ \cdot y^{2}

Combine the Results

Finally, combine the simplified terms outside the root while keeping any terms that are not perfect powers inside the radical.

2 \cdot 5 \cdot ∣ x^{3} ∣ \cdot y^{2} = 52 ∣ x^{3} ∣ y^{2}

Note that, depending on how complex the given radical expression is, different techniques, such as rationalizing or combining radicals, can be applied.

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Proof

a≥0

a<0

Even n

Odd n

Summary

Extra

Proof

Proof

Proof

$a \geq 0$

$a < 0$

Even $n$

Odd $n$