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 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.
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.