# 5. Multiplication and Division of Radicals

### Later, on this page...

When multiplying expressions containing radicals, we use the following law, along with normal procedures of algebraic multiplication.

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

### Example 1

(a) `sqrt5sqrt2`

Answer

`sqrt(5)sqrt(2) = sqrt (5 times 2) = sqrt(10)`

Nothing much to do here - since both items involve a square root, we can combine them by multiplying the radicands.

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(b) `sqrt33sqrt3`

Answer

`sqrt(33)sqrt(3) `

`= sqrt(99)`

` = sqrt(9 times 11) `

`= 3sqrt(11)`

After multiplying the radicands, we can see that there is a square number under the square root sign (the 9), for which we can find the square root.

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### Example 2

(a) `root(3)6root(3)4`

Answer

`root(3)(6) root(3)(4)`

` = root(3)(6 times 4)`

` = root(3)(24)`

`=root(3)(8 times 3)`

` = root(3)(8)root(3)(3)`

` = 2 root(3)(3)`

In this case, we needed to find the largest cube that divides into `24`, and the answer was `8`.

(The "cubes" are the numbers `1^3= 1`, `2^3= 8`, `3^3= 27`, `4^3= 64`, ...)

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(b) `root(5)(8a^3b^4)root(5)(8a^2b^3)`

Answer

Our aim here is to recognize and factor out terms which are 5th powers, so that they can be simplified using the 5th-root.

`root(5)(8a^3b^4) root(5)(8a^2b^3)`

We first multiply out the expressions under the 5th roots.

` = root(5)((8a^3b^4)(8a^2b^3))`

`= root(5)(64a^5b^7)`

Then we look for numbers or algebraic terms for which we can find the 5th root.

`=root(5)(32a^5b^5) root(5)(2b^2)`

Then we find those fifth roots, and leave the `2b^2` under the 5th root sign because we cannot do any more with it.

`=2ab root(5)(2b^2) `

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### Example 3

(a) `(3+sqrt5)^2`

Answer

Recall that in general,

`(a+b)^2=a^2+2ab+b^2`

We use this to expand out our given expression.

`(3 + sqrt(5))^2`

` = 3^2 + 2(3)(sqrt(5)) + (sqrt(5))^2`

`= 9 + 6sqrt(5) + 5`

`= 14 + 6sqrt(5)`

(b) `(sqrta-sqrtb)^2`

Answer

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

` = (sqrt(a))^2 - 2(sqrt(a))(sqrt(b)) + (sqrt(b))^2`

`=a + b - 2 sqrt(ab)`

(c) `(5+sqrta)(5-sqrta)`

Answer

We recognize that we have an expression in the form (*x* + *y*)(*x* − *y* and recall from earlier algebra that this is equal to the difference of 2 squares.

(

x+y)(x−y) =x^{2}−y^{2}

We use this to find our answer as follows:

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

Notice that our final answer does not have a square root sign. We now havea **rational** answer (since there are no radicals in it).

This idea is important in the next section, where we use it to rationalize the denominator of a fraction.

## Division of Radicals (Rationalizing the Denominator)

This process is also called "rationalising the denominator" since we remove all irrational numbers in the denominator of the fraction.

This is important later when we come across Complex Numbers.

**Reminder:** From earlier algebra, you will recall the difference of squares formula:

(

a+b)(a−b) =a^{2}−b^{2}

We will use this formula to rationalize denominators.

### Example 4

Simplify: `1/(sqrt3-sqrt2`

Answer

The question requires us to divide 1 by (√3 − √2).

We need to multiply top and bottom of the fraction by the **conjugate** of (√3 − √2).

The conjugate is easily found by reversing the sign in the middle of the radical expression. In this case, our minus becomes plus. So the conjugate of (√3 − √2) is (√3 + √2).

`frac{1}{sqrt3 - sqrt2}`

` = frac{1}{sqrt3 - sqrt2} times frac{sqrt3 + sqrt 2}{sqrt3 + sqrt2}`

`= frac{sqrt3 + sqrt2}{(sqrt3)^2 - (sqrt(2))^2}`

`=frac{sqrt3 + sqrt2}{3-2}`

`=sqrt3 + sqrt2`

After we multiply top and bottom by the conjugate, we see that the denominator becomes free of radicals (in this case, the denominator has value 1).

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### Historical Note

In the days before calculators, it was important to be able to rationalize denominators. Using logarithm tables, it was very troublesome to find the value of expressions like our example above.

Now that we use calculators, it is not so important to rationalize denominators.

However, rationalizing denominators is still used for several of our algebraic techniques (see especially Complex Numbers), and is still worth learning.

### Exercises

Q1 `root(5)4root(5)16`

Answer

`root(5)(4) root(5)(16)`

`=root(5)(4 times 16)`

`=root(5)(64)`

`=root(5)(32 times 2)`

`= 2 root(5)(2)`

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Q2 `sqrta(sqrt(ab)+sqrt(c^3))`

Answer

`sqrta(sqrt(ab) + sqrt(c^3))`

` = sqrt(aab) + sqrt(ac^3)`

`= sqrt(a^2 b)+ sqrt(ac c^2)`

`=asqrtb + csqrt(ac)`

Q3. `(2sqrtx)/(sqrtx-sqrty)`

Answer

The conjugate of `(sqrtx - sqrty)` is `(sqrtx + sqrty)`. We multiply top and bottom of the fraction by this conjugate to remove the surds in the denominator.

`frac{2sqrtx}{sqrtx - sqrty} `

`= frac{2sqrtx}{sqrtx - sqrty} times frac{sqrtx + sqrty}{sqrtx + sqrty}`

`=frac{2x+2sqrt(xy)}{x-y}`

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Q4. `(6sqrta)/(2sqrta-b)`

Answer

The conjugate of `(2sqrta - b)` is `(2sqrta + b)` and we multiply top and bottom of the fraction by this conjugate.

`frac{6sqrta}{2sqrta - b}`

`=frac{6sqrta}{2sqrta - b} times frac{2sqrta + b}{2sqrta + b}`

`=frac{12a+6bsqrta}{4a - b^2}`

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