Before we can simplify radicals, we need to know some rules about them. These rules just follow on from what we learned in the first 2 sections in this chapter, Integral Exponents and Fractional Exponents.

Expressing in simplest radical form just means simplifying a radical so that there are no more square roots, cube roots, 4th roots, etc left to find. It also means removing any radicals in the denominator of a fraction.

Let's take the positive case first.

### n-th root of a Positive Number to the Power n

We met this idea in the last section, Fractional Exponents. Basically, finding the n-th root of a (positive) number is the opposite of raising the number to the power n, so they effectively cancel each other out. These 4 expressions have the same value:

root(n)(a^n)=(root(n)a)^n=root(n)((a^n))=a

The 2nd item in the equality above means:

"take the n-th root first, then raise the result to the power n"

The 3rd item means:

"raise a to the power n then find the n-th root of the result"

Both steps lead back to the a that we started with.

For the simple case where n = 2, the following 4 expressions all have the same value:

sqrt(a^2)=(sqrt(a))^2=sqrt((a^2))=a

For example, if a = 9:

sqrt(9^2)=(sqrt(9))^2=sqrt((9^2))=9

The second item means: "Find the square root of 9 (answer: 3) then square it (answer 9)".

The 3rd item means: "Square 9 first (we get 81) then find the square root of the result (answer 9)".

In general we could write all this using fractional exponents as follows:

root(n)(a^n)=(a^(1//n))^n=(a^n)^(1//n)=a

Yet another way of thinking about it is as follows:

(a^(1/n))^n=a^((1/nxxn))=a

### n-th root of a Negative Number to the Power n

We now consider the above square root example if the number a is negative.

For example, if a = -5, then:

sqrt((-5)^2)=sqrt(25)=5

A negative number squared is positive, and the square root of a positive number is positive.

In general, we write for a, a negative number:

sqrt((a)^2)=|a|

Notice I haven't included this part: (sqrt(a))^2. In this case, we would have the square root of a negative number, and that behaves quite differently, as you'll learn in the Complex Numbers chapter later.

Continues below

### The Product of the n-th root of a and the n-th root of b is the n-th root of ab

root(n)axxroot(n)b=root(n)(ab)

Example:

root(4)7xxroot(4)5=root(4)(7xx5)=root(4)35

We could write "the product of the n-th root of a and the n-th root of b is the n-th root of ab" using fractional exponents as well:

a^(1//n)xxb^(1//n)=(ab)^(1//n)

### The m-th Root of the n-th Root of the Number a is the mn-th Root of a

root(m)(root(n)a)=root(m\ n)a

We could write this as:

(a^(1//n))^(1//m)=(a)^(1//(mn))

Example:

root(4)(root(3)5)=root(12)5

This has the same meaning:

(5^(1//3))^(1//4)=(5)^(1//(12))

In words, we would say: "The 4th root of the 3rd root of 5 is equal to the 12th root of 5".

### The n-th Root of a Over the n-th Root of b is the n-th Root of a/b

root(n)a/root(n)b=root(n)(a/b)(b ≠ 0)

Example:

root(3)375/root(3)3=root(3)(375/3)=root(3)125=5

If we write the our general expression using fractional exponents, we have:

a^(1//n)/b^(1//n)=(a/b)^(1//n) (b ≠ 0)

### Mixed Examples

Simplify the following:

(a) root(5)(4^5)

root(5)(4^5)=(root(5)4)^5=4

We have used the first law above,

root(n)(a^n)=(a^(1//n))^n=(a^n)^(1//n)=a

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(b) root(3)2root3(3)

root(3)2root3(3)=root(3)(2xx3)=root(3)6

We have used the rule:

a^(1//n)xxb^(1//n)=(ab)^(1//n).

(c) root(3)sqrt5

root(3)sqrt5=root(3xx2)5=root(6)5

We have used the law: (a^(1//n))^(1//m)=a^(1//mn)

(d) sqrt7/sqrt3

sqrt7/sqrt3=sqrt(7/3)

Nothing much to do here. We used: a^(1//n)/b^(1//n)=(a/b)^(1//n)

In these examples, we are expressing the answers in simplest radical form, using the laws given above.

(a) sqrt72

We need to examine 72 and find the highest square number that divides into 72. (Squares are the numbers 1^2= 1,   2^2= 4,   3^2= 9,   4^2= 16, ...)

In this case, 36 is the highest square that divides into 72 evenly. We express 72 as 36 × 2 and proceed as follows.

sqrt72=sqrt(36xx2)=sqrt(36)sqrt(2)=6sqrt(2)

We have used the law: a^(1//n)xxb^(1//n)=(ab)^(1//n)

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(b) sqrt(a^3b^2)

sqrt(a^3b^2)

=sqrt(a^2xxaxxb^2)

=sqrt(a^2)xxsqrt(a)xxsqrt(b^2)

=ab sqrt(a)

We have used the law: sqrt(a^2)=a.

(c) root(3)40

root(3)40 = root(3)(8xx5) = root(3)8 xxroot(3) 5= 2 root(3)5

(d) root(5)(64x^8y^(12))

root(5)(64x^8y^(12))

=root(5)(32xx2xxx^5xxx^3xxy^10xxy^2)

=root(5)(32x^5y^10) root(5)(2x^3y^2)

=2xy^2root(5)(2x^3y^2)

### Exercises

Simplify:

Q1 sqrt(12ab^2)

sqrt(12ab^2)

= sqrt(3xx4ab^2)

=sqrt(3)sqrt(4)sqrt(a)sqrt(b^2)

=2bsqrt(3a)

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Q2 root(4)(64r^3s^4t^5

root(4)(64r^3s^4t^5)

We factor out all the terms that are 4th power. The number 16 is a 4th power, since 2^4= 16.

=root(4)((16xx4)r^3s^4(t^4xxt))

=root(4)(16s^4t^4)xx(root(4)(4r^3t))

=root(4)(2^4s^4t^4)xx(root(4)(4r^3t))

=root(4)(2^4)xxroot(4)(s^4)xxroot(4)(t^4)xx(root(4)(4r^3t))

Then we find the 4th root of each of those terms.

=2stroot(4)(4r^3t)

There are no 4th powers left in the expression 4r^3t, so we leave it under the 4th root sign.

Q3 sqrt(x/(2x+1)

This one requires a special trick. To remove the radical in the denominator, we need to multiply top and bottom of the fraction by the denominator.

sqrt(x/(2x+1)

=sqrtx/(sqrt(2x+1))xx(sqrt(2x+1))/(sqrt(2x+1))

=(sqrt(x)sqrt(2x+1))/(2x+1)

We can see that the denominator no longer has a radical.

In the days before calculators, it was important to be able to rationalise a denominator like this. You can see more examples of this process in 5. Muliplication and Division of Radicals.

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