# How do you rationalize the denominator and simplify #(2sqrt7+sqrt3) / (3sqrt7-sqrt3)#?

##### 1 Answer

#### Explanation:

Your starting expression is

#(2sqrt(7) + sqrt(3))/(3sqrt(7) - sqrt(3))#

To rationalize the denominator, you need to multiply it by its *conjugate*. For any binomial, you get its conjugate by *negating* the second term.

In essence, you need to change the sign of the second term.

In your case, that would imply

#underbrace(3sqrt(7) - sqrt(3))_(color(blue)("original binomial")) -> overbrace(3sqrt(7) + sqrt(3))^(color(red)("its conjugate"))#

So, multiply the fraction by

#(2sqrt(7) + sqrt(3))/(3sqrt(7) - sqrt(3)) * (3sqrt(7) + sqrt(3))/(3sqrt(7) + sqrt(3)) = ( (2sqrt(7) + 3)(3sqrt(7) + 3))/((3sqrt(7) - 3)(3sqrt(7) + 3))#

The denominator of the fraction is now in the form

#color(blue)((a-b)(a+b) = a^2 - b^2)#

which means that you can write it as

#(3sqrt(7) - 3)(3sqrt(7) + 3) = (3sqrt(7))^2 - 3^2#

#=9 * 7 - 9 = 54#

For the numerator, expand the parantheses to get

#(2sqrt(7) + 3)(3sqrt(7) + 3) = 2sqrt(7) * 3sqrt(7) + 6sqrt(7) + 9sqrt(7) + 9#

#=6 * 7 + 15sqrt(7) + 9 = 51 + 15sqrt(7)#

The fraction will now be

#( (2sqrt(7) + 3)(3sqrt(7) + 3))/((3sqrt(7) - 3)(3sqrt(7) + 3)) = (51 + 15sqrt(7))/54#

Finally, simplify to get

#(51 + 15sqrt(7))/54 = color(green)((17 + 5sqrt(7))/18)#