# Given that `(2-3sqrt(7))/(2-3sqrt(7))=a+bsqrt(7)` , find a and b.

ishpiro | Certified Educator

Parts of my answer seem to have disappeared. This is what missing from the beginning:

In the given fraction, either numerator or denominator has to contain "plus" instead of "minus", otherwise it would just equal 1. Assuming there is a "plus" in the numerator, we can rationalize the denominator by multiplying both top and bottom by `2+3sqrt(7)` .

This is what missing from the end:

This can also be written as

`(67+12sqrt(7))/(-59) = -67/59-12/59sqrt(7)`

So these are the values for a and b:

`a=-67/59` , `b=-12/59`

If in the original fraction there was a "minus" on the top and "plus" on the bottom, we would multiply top and bottom by `2-3sqrt(7)`

and then the answer would be

`a=-67/59` , `b=12/59`

ishpiro | Certified Educator

`2-3sqrt(7)` and `2+3sqrt(7)` are called "conjugates" of each other. So now we need to simplify

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

First, simplify the numerator by multiplying the two radical expressions using FOIL:

`(2+3sqrt(7))(2+3sqrt(7)) = 4+6sqrt(7)+6sqrt(7) +9(sqrt(7))^2 = 4+12sqrt(7) +9*7 = 67+12sqrt(7)`

Now, simplify the denominator by also using FOIL. Notice that inner and outer terms, both containing radicals, will cancel each other out:

`(2-3sqrt(7))(2+3sqrt(7)) = 4+6sqrt(7)-6sqrt(7)-9(sqrt(7))^2 = 4-9*7=-59`

Putting this back into the fraction, we have

`(2+3sqrt(7))/(2-3sqrt(7)) * (2+3sqrt(7))/(2+3sqrt(7)) = (67+12sqrt(7))/(-59)`

This can also be written as

ishpiro | Certified Educator

It would have to be a = 8 and `b=-sqrt(7)` .

It is possible. But it was more likely that the intended question was to rationalize the denominator and a and b were supposed to be rational numbers. Good point though :)

alexeyhurricane | Student

Why "a" can't be -8 and b sqrt(7) if left hand side =1