# 3. Simplest Radical Form

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.

## Laws of Radicals

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 powern"

The 3rd item means:

"raise

ato the powernthen find then-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.

### 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)`

Answer

`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`

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

Answer

`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`

Answer

`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`

Answer

`sqrt7/sqrt3=sqrt(7/3)`

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

### Simplest Radical Form Examples

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

(a) `sqrt72`

Answer

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)`

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

Answer

`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`

Answer

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

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

Answer

`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)`

Answer

`sqrt(12ab^2) `

`= sqrt(3xx4ab^2)`

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

`=2bsqrt(3a)`

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

Answer

`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.

Answer

`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.