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Properties of Radicals
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Reference
Properties of Radicals
Rule
n^\text{th} Roots of n^\text{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 sqrt(a^n), the radical expression can be simplified as follows.
sqrt(a^n)= a, & if n is odd |a|, & 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^\text{th} roots that can be simplified by using the formula.
| a | n | sqrt(a^n) | Is n even? | Simplify |
|---|---|---|---|---|
| 4 | 3 | sqrt(4^3) | * | 4 |
| -6 | 5 | sqrt((-6)^5) | * | -6 |
| 2 | 6 | sqrt(2^6) | ✓ | |2| = 2 |
| -3 | 4 | sqrt((-3)^4) | ✓ | |-3| = 3 |
Proof
Informal JustificationTo write the expression for sqrt(a^n), there are two cases to consider.
- a≥ 0
- a<0
Both cases will be considered one at a time.
a≥ 0
Start by noting that since a is non-negative, a^n is also non-negative. This means that the n^\text{th} root can be rewritten using a rational exponent. sqrt(a^n) = ( a^n )^(1n) Since both n and 1n are rational numbers, the Power of a Power Property for Rational Exponents can be applied to simplify the obtained expression.
Therefore, sqrt(a^n) is equal to a if 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 sqrt(a^(n_e)).
sqrt(a^(n_e))
â–¼
Simplify
NegBaseToPosPow
(- a)^(n_e) = a^(n_e)
sqrt((- a)^(n_e))
RootToPowD
sqrt(a)=a^(1n)
( (- a)^(n_e) )^()1n_e
PowPow
(a^m)^n=a^(m* n)
(- a)^(n_e* 1n_e)
a * 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^\text{th} root, the expression sqrt(a^(n_o)) is the number y that, when multiplied by itself n_o times, will result in a^(n_o). y * y * ... * y_(n_otimes)=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. sqrt(a^n) = a
Summary
If a is non-negative, sqrt(a^n) is always equal to a. However, in case of negative a, the value of sqrt(a^n) depends on the parity of n.
| a≥ 0 | a<0 | |
|---|---|---|
| Even n | sqrt(a^n) = a | sqrt(a^n) = - a |
| Odd n | sqrt(a^n) = a | sqrt(a^n) = a |
To conclude, for odd values of n, the expression sqrt(a^n) is equal to a. On the other hand, if n is even, sqrt(a^n) can be written as |a|.
sqrt(a^n)= a, & if n is odd |a|, & if n is even
Extra
Simplifying sqrt(a^m)Depending on the index of the root and the power in the radicand, simplifying sqrt(a^m) may be problematic. Because real n^\text{th} roots with an even index are defined only for non-negative numbers, the absolute value is sometimes needed.
If n is even, sqrt(a^m) is defined only for non-negative a^m.
The following applet presents a decision tree to simplify sqrt(a^m). In this applet, k is the greatest common factor of m and n.
Consider example n^\text{th} roots sqrt(a^m) that can be simplified by using the decision tree.
| a | m | n | sqrt(a^m) | Simplify |
|---|---|---|---|---|
| 2 | 8 | 4 | sqrt(2^8) | 2^()84 = 2^2 = 4 |
| -3 | 6 | 2 | sqrt((-3)^6) | | (-3)^()62 | = | (-3)^3 | = |-27| = 27 |
| x | 6 | 3 | sqrt(x^6) | x^()63 = x^2 |
| -3 | 2 | 8 | sqrt((-3)^2) | sqrt(|-3|) = sqrt(3) |
| 2 | 3 | 6 | sqrt(2^3) | sqrt(2) |
| -2 | 3 | 9 | sqrt((-2)^3) | sqrt(-2) |
| 2 | 6 | 8 | sqrt(2^6) | sqrt(2^3) = sqrt(8) |
| -2 | 6 | 8 | sqrt((-2)^6) | sqrt(|-2|^3) = sqrt(2^3) = sqrt(8) |
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Rule
Product Property of Radicals
Given two non-negative numbers a and b, the n^\text{th} root of their product equals the product of the n^\text{th} root of each number.
sqrt(ab) = sqrt(a)* sqrt(b), for a≥ 0 and b≥ 0
If n is an odd number, the n^\text{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 or b is equal to 0. By the Zero Property of Multiplication, the radicand in left-hand side of the formula is equal to 0.
\begin{array}{ccc}
\underline\textbf{Left-Hand Side} & & \underline\textbf{Right-Hand Side} \\[0.5em]
\sqrt[n]{0} & \stackrel{?}{=} & \sqrt[n]{a}\cdot \sqrt[n]{b}
\end{array}
Because 0^n = 0, n^\text{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.
\begin{array}{ccc}
\underline\textbf{Left-Hand Side} & & \underline\textbf{Right-Hand Side} \\[0.5em]
0 & = & 0
\end{array}
Therefore, the property is true when a or 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 = sqrt(a), y = sqrt(b), and z = sqrt(ab). By the definition of an n^\text{th} root, each of these numbers raised to the n^\text{th} power is equal to its corresponding radicand.
x^n = a & (I) y^n = b & (II) z^n = ab & (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
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Substitute values and simplify
Substitute
y^n= b
x^n y^n = a b
Substitute
ab= z^n
x^n y^n = z^n
ProdPowII
a^m b^m = (a b)^m
(xy)^n = z^n
RearrangeEqn
Rearrange equation
z^n = (xy)^n
Since xy and z are of the same sign, the final equation implies that z=xy. z^n=(xy)^n ⇒ z=xy The last step is substituting z=sqrt(ab), x=sqrt(a), and y=sqrt(b) into this equation. z=xy ⇔ sqrt(ab)=sqrt(a)* sqrt(b)
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Rule
Quotient Property of Radicals
Let a be a non-negative number and b be a positive number. The n^\text{th} root of the quotient ab equals the quotient of the n^\text{th} roots of a and b.
sqrt(a/b) = sqrt(a)/sqrt(b), for a≥ 0, b > 0
If n is an odd number, the n^\text{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 0b=0, the radicand in left-hand side of the formula is equal to 0.
\begin{array}{ccc}
\underline\textbf{Left-Hand Side} & & \underline\textbf{Right-Hand Side} \\[0.5em]
\sqrt[n]{0} & \stackrel{?}{=} & \dfrac{\sqrt[n]{0}}{\sqrt[n]{b}}
\end{array}
Because 0^n = 0, n^\text{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.
\begin{array}{ccc}
\underline\textbf{Left-Hand Side} & & \underline\textbf{Right-Hand Side} \\[0.5em]
0 & \stackrel{{\color{#009600}{\checkmark}}}{=} & 0
\end{array}
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 = sqrt(a), y = sqrt(b), and z = sqrt(ab). By the definition of an n^\text{th} root, each of these numbers raised to the n^\text{th} power is equal to its corresponding radicand.
x^n = a & (I) y^n = b & (II) z^n = ab & (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 ⇓ x^n/y^n = a/y^n
Now, substitute Equations (II) and (III) into the obtained equation.
x^n/y^n = a/y^n
â–¼
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
a^m/b^m=(a/b)^m
(x/y)^n = z^n
RearrangeEqn
Rearrange equation
z^n = (x/y)^n
Since xy and z are of the same sign, the final equation implies that z= xy. z^n=(x/y)^n ⇒ z=x/y The last step is substituting z=sqrt(ab), x=sqrt(a), and y=sqrt(b) into this equation. z=x/y ⇔ sqrt(a/b)=sqrt(a)/sqrt(b)
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Rule
Quotient Property of Radicals
Let a be a non-negative number and b be a positive number. The n^\text{th} root of the quotient ab equals the quotient of the n^\text{th} roots of a and b.
sqrt(a/b) = sqrt(a)/sqrt(b), for a≥ 0, b > 0
If n is an odd number, the n^\text{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 0b=0, the radicand in left-hand side of the formula is equal to 0.
\begin{array}{ccc}
\underline\textbf{Left-Hand Side} & & \underline\textbf{Right-Hand Side} \\[0.5em]
\sqrt[n]{0} & \stackrel{?}{=} & \dfrac{\sqrt[n]{0}}{\sqrt[n]{b}}
\end{array}
Because 0^n = 0, n^\text{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.
\begin{array}{ccc}
\underline\textbf{Left-Hand Side} & & \underline\textbf{Right-Hand Side} \\[0.5em]
0 & \stackrel{{\color{#009600}{\checkmark}}}{=} & 0
\end{array}
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 = sqrt(a), y = sqrt(b), and z = sqrt(ab). By the definition of an n^\text{th} root, each of these numbers raised to the n^\text{th} power is equal to its corresponding radicand.
x^n = a & (I) y^n = b & (II) z^n = ab & (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 ⇓ x^n/y^n = a/y^n
Now, substitute Equations (II) and (III) into the obtained equation.
x^n/y^n = a/y^n
â–¼
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
a^m/b^m=(a/b)^m
(x/y)^n = z^n
RearrangeEqn
Rearrange equation
z^n = (x/y)^n
Since xy and z are of the same sign, the final equation implies that z= xy. z^n=(x/y)^n ⇒ z=x/y The last step is substituting z=sqrt(ab), x=sqrt(a), and y=sqrt(b) into this equation. z=x/y ⇔ sqrt(a/b)=sqrt(a)/sqrt(b)
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Method
Simplifying Radical Expressions
Expressions involving radicals be simplified using various properties of exponents and radicals. Consider the following square root. sqrt(50x^6y^4) Follow these four steps to simplify the expression.
1
Factor the Radicand
expand_more 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 * 5^2
Since x^6 and y^4 are already prime factors, the radical expression can be rewritten as follows.
sqrt(50x^6y^4) = sqrt(2 * 5^2 * x^6 * y^4)
2
Separate Perfect Powers
expand_more 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.
sqrt(2 * 5^2 * x^6 * y^4) = sqrt(2 * 5^2 * (x^3)^2 * (y^2)^2)
3
Simplify the Perfect Powers
expand_more 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.
sqrt(2 * 5^2 * (x^3)^2 * (y^2)^2)
SqrtProd
sqrt(a* b)=sqrt(a)*sqrt(b)
sqrt(2) * sqrt(5^2) * sqrt((x^3)^2) * sqrt((y^2)^2)
SqrtPowToNumber
sqrt(a^2)=a
sqrt(2) * 5 * sqrt((x^3)^2) * sqrt((y^2)^2)
SqrtToAbs
sqrt(a^2)=|a|
sqrt(2) * 5 * |x^3| * |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. sqrt(2) * 5 * |x^3| * |y^2| =sqrt(2) * 5 * |x^3| * y^2
4
Combine the Results
expand_more Finally, combine the simplified terms outside the root while keeping any terms that are not perfect powers inside the radical.
sqrt(2) * 5 * |x^3| * y^2 = 5sqrt(2)|x^3| y^2
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