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# Multiplying Radicals â€“ Techniques & Examples

A radical can be defined as a symbol that indicates the root of a number. Square root, cube root, fourth root are all radicals.

Mathematically, a radical is represented as x ^{n}. This expression tells us that a number x is multiplied by itself n number of times.

## How to Multiply Radicals?

For example, the multiplication of âˆša with âˆšb is written as âˆša x âˆšb. Similarly, the multiplication n ^{1/3}Â with y ^{1/2}Â is written as h ^{1/3}y ^{1/2}.

It advisable to place factors in the same radical sign. This is possible when the variables are simplified to a common index. For example, the multiplication of ^{n}âˆšx with ^{n }âˆšyÂ is equal toÂ ^{n}âˆš(xy). This means that the root of several variables’ product is equal to the product of their roots.

*Example 1*

Multiply âˆš8xbÂ by âˆš2xb.

__Solution__

âˆš8xbÂ by âˆš2xb = âˆš(16x ^{2 }b ^{2})Â = 4xb.

You can notice that the multiplication of radical quantities results in rational quantities.

*Â *

*Example 2*

Find the product of âˆš2Â and âˆš18.

__Solution__

âˆš2Â x âˆš18 = âˆš36Â = 6.

### Multiplication of Quantities when the Radicands are of the Same Value

Roots of the same quantity can be multiplied by the addition of the fractional exponents. In general,

a ^{1/2} * a ^{1/3}Â = a ^{(1/2 + 1/3)}Â = a ^{5/6}

In this case, the denominator’s sum indicates the root of the quantity, whereas the numerator denotes how the root is to be repeated to produce the required product.

### Multiplication of Radical Quantities with Rational Coefficients

The radicals’ rational parts are multiplied, and their product is prefixed to the product of the radical quantities. For instance, aâˆšb x câˆšd = ac âˆš(bd).

*Example 3*

Find the following product:

âˆš12x * âˆš8xy

__Solution__

- Multiply all quantities the outside of radical and all quantities inside the radical.

âˆš96x ^{2 }y

- Simplify the radicals

4xâˆš6 y

*Example 4*

Solve the following radical expression

(3 + âˆš5)/(3Â – âˆš5) +Â (3 – âˆš5)/(3Â + âˆš5)

__Solution__

- Find the LCM to get,

[(3 +âˆš5)Â² + (3-âˆš5)Â²]/[(3+âˆš5)(3-âˆš5)]

- ExpandÂ (3 + âˆš5) Â² and (3 – âˆš5) Â² as,

3 Â² + 2(3)(âˆš5) + âˆš5 Â² and 3 Â²- 2(3)(âˆš5) + âˆš5 Â² respectively.

- Add the above two expansions to find the numerator,

3 Â² + 2(3)(âˆš5) + âˆš5 Â² + 3 Â² – 2(3)(âˆš5) + âˆš5 Â² = 18 + 10 = 28

- Compare the denominator (3-âˆš5)(3+âˆš5)Â with identity a Â² – b Â²= (a + b)(a – b), to get

3 Â² – âˆš5 Â² = 4

- Write the final answer,

28/4 = 7

*Example 5*

Rationalize the denominator [(âˆš5 – âˆš7)/(âˆš5 + âˆš7)] – [(âˆš5 + âˆš7) / (âˆš5Â – âˆš7)]

__Solution__

- By calculating the L.C.M, we get

(âˆš5 – âˆš7) Â² – (âˆš5 + âˆš7) Â² / (âˆš5 + âˆš7)(âˆš5 – âˆš7)

- Expansion ofÂ (âˆš5Â – âˆš7) Â²

= âˆš5 Â² + 2(âˆš5)(âˆš7) +Â âˆš7Â²

- Expansion ofÂ (âˆš5 + âˆš7) Â²

= âˆš5 Â² – 2(âˆš5)(âˆš7) +Â âˆš7 Â²

- Compare the denominatorÂ (âˆš5 +Â âˆš7)(âˆš5Â – âˆš7)Â with the identity aÂ² – b Â²Â =Â (a + b)(a – b), to get,

âˆš5 Â²Â – âˆš7 Â² = -2

- Solve,

[{âˆš5 Â² + 2(âˆš5)(âˆš7) +Â âˆš7Â²} â€“ {âˆš5 Â² – 2(âˆš5)(âˆš7) +Â âˆš7 Â²}]/(-2)

= 2âˆš35/(-2)

= -âˆš35

*Â *

*Example 6*

Evaluate

(2 + âˆš3)/(2 – âˆš3)

__Solution__

- In this case, 2 – âˆš3 is the denominator and rationalizes the denominator, both top, and bottom by its conjugate.

The conjugate ofÂ 2 – âˆš3 isÂ 2 + âˆš3.

- Comparing the numerator (2 + âˆš3) Â²Â with the identityÂ (a + b) Â²= a Â²+ 2ab + b Â², the result is 2 Â² + 2(2)âˆš3 +Â âˆš3Â² =Â Â (7 + 4âˆš3).
- Comparing the denominator with the identityÂ (a + b) (a – b) =Â a Â² – b Â², the results is 2Â² –Â âˆš3Â².
- Answer = (7 + 4âˆš3)

*Example 7*

Multiply âˆš27/2 x âˆš(1/108)

__Solution__

âˆš27/2 x âˆš(1/108)

= âˆš27/âˆš4 x âˆš(1/108)

= âˆš(27 / 4) x âˆš(1/108)

= âˆš(27 / 4) x âˆš(1/108) = âˆš(27 / 4 x 1/108)

= âˆš(27 / 4 x 108)

Since 108 = 9 x 12 and 27 = 3 x 9

âˆš(3 x 9/ 4 x 9 x 12)

9 is a factor of 9, and so simplify,

âˆš(3 / 4 x 12)

= âˆš(3 / 4 x 3 x 4)

= âˆš(1 / 4 x 4)

=âˆš(1 / 4 x 4) = 1 / 4