Question

Question #3f390

Answer 1

`-2/3sqrt(3)-sqrt(2)`

Explanation:

Consider the just the denominator for a moment.

Demonstrating a principle by example:

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

Now compare to:

`(sqrt(12))^2-(sqrt(18))^2 = (sqrt(12)-sqrt(18))(sqrt(12)+sqrt(18))`

So to remove the roots from the denominator we need to 'force' it to be `(sqrt(12)-sqrt(18))(sqrt(12)+sqrt(18))`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Getting rid of the roots in the denominator")`

Multiply by 1 and you do not change the value. However, 1 comes in many forms so you can change the way numbers look without changing their intrinsic value.

`color(green)(2/(sqrt(12)-sqrt(18))color(red)(xx1))`

`color(green)(2/(sqrt(12)-sqrt(18))color(red)( xx(sqrt(12)+sqrt(18))/(sqrt(12+sqrt(18))`

`(2(sqrt(12)+sqrt(18)))/( (sqrt(12))^2-(sqrt(18))^2)`

`(2(sqrt(12)+sqrt(18)))/( 12-18)`

`(2(sqrt(12)+sqrt(18)))/( -6)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Simplifying")`

`(cancel(2)^1(sqrt(12)+sqrt(18)))/( -cancel(6)^3)`

`-1/3(sqrt(12)+sqrt(18))`

But `sqrt(12) = sqrt(3xx2^2)=2sqrt(3)`
And `sqrt(18)=sqrt(2xx3^2)=3sqrt(2)`

`-1/3(2sqrt(3)+3sqrt(2))`

`-2/3sqrt(3)-sqrt(2)`

Answer 2

`-2/3sqrt3-sqrt2`

Explanation:

Multiply the numerator/denominator of the fraction by the `color(blue)"conjugate"` of the denominator.

The `color(blue)"conjugate"" of "sqrt12-sqrt18" is "sqrt12color(red)(+)sqrt18`

Considering the denominator.

`color(orange)"Reminder " (sqrtaxxsqrta)=a`

`"Expand " (sqrt12-sqrt18)(sqrt12color(red)(+)sqrt18)` using the FOIL method.

`=12+cancel(sqrt18xxsqrt12)cancel(-sqrt18xxsqrt12)-18`

`=-6larrcolor(red)" rational denominator"`

`rArr2/(sqrt12-sqrt18)`

`=2/(sqrt12-sqrt18)xx(sqrt12+sqrt18)/(sqrt12+sqrt18)`

`=(2(sqrt12+sqrt18))/(-6)`

`=-1/3(sqrt12+sqrt18)`

Simplifying `sqrt12" and " sqrt18`

`sqrt12=sqrt(4xx3)=sqrt4xxsqrt3=2sqrt3`

`sqrt18=sqrt(9xx2)=sqrt9xxsqrt2=3sqrt2`

`rArr-1/3(sqrt12+sqrt18)`

`=-1/3(2sqrt3+3sqrt2)larr" this is acceptable"`

`=-2/3sqrt3-sqrt2`