Before we have the right to simplify radicals, we require to know some rules about them. These rules just follow on from what us learned in the an initial 2 part in this chapter, Integral Exponents and Fractional Exponents.

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Expressing in simplest radical form just way simplifying a radical so that there space no an ext square roots, cube roots, 4th roots, and so on left come find. That also method removing any kind of radicals in the denominator of a fraction.

Let"s take the positive instance first.

### n-th root of a positive Number come the strength n

We met this idea in the last section, fractional Exponents.Basically, finding the n-th root of a (positive) number is the opposite ofraising the number to the power n, therefore they effectively cancel eachother out. These 4 expressions have actually the very same value:

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

The second item in the equality above means:

"take the n-th source first, climate raise the an outcome to the power n"

The 3rd item means:

"raise a to the strength n then find the n-th source of the result"

Both actions lead earlier to the a the we started with.

For the an easy case where `n = 2`, the complying with 4 expression all have the exact 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 2nd item means: "Find the square root of `9` (answer: `3`) then square it (answer `9`)".

The third item means: "Square `9` an initial (we gain `81`) then uncover the square root of the result (answer `9`)".

In general we can write every this making use of fractional exponents as follows:

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

Yet another method of thinking around it is as follows:

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

### n-th source of a an adverse Number come the strength n

We now take into consideration the above square root example if the number `a` is negative.

For example, if `a = -5`, then:

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

A an adverse number squared is positive, and also the square root of a optimistic 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 source of a an adverse number, and that behaves rather differently, together you"ll learn in the complicated Numbers thing later.

### The Product the the n-th root of a and also the n-throot 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 that the n-th source of a and the n-throot the b is the n-th source of ab" utilizing fractional exponents as well:

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

### The m-th source of the n-th root of the Number ais the mn-th source of a

`root(m)(root(n)a)=root(m n)a`

We might write this as:

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

Example:

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

This has actually the same meaning:

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

In words, we would say: "The fourth root that the third root the `5` is equal to the 12th root of `5`".

### The n-th source of a over the n-thRoot that 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, us have:

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

### Mixed instances

Simplify the following:

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

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

We have used the very first law above,

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

In these examples, we room expressing the answers in easiest radical form, making use of the laws provided above.

(a) `sqrt72`

We must examine `72` and also find the greatest square number that divides right into `72`. (Squares space the numbers `1^2= 1`, `2^2= 4`, `3^2= 9`, `4^2= 16`, ...)

In this case, `36` is the greatest square the divides right into `72` evenly. We express `72` together `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)`

`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`.

`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)`

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

We element out every the terms that are fourth 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 uncover the 4th root of every of those terms.

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

There space no 4th powers left in the expression `4r^3t`, so us leave that under the 4th root sign.

Q3 `sqrt(x/(2x+1)`

This one requires a unique trick. To eliminate the radical in the denominator, we should multiply top and bottom that the portion by the denominator.