How do you rationalize #(sqrt6 - 3 ) / 4#?

1 Answer
George C.
May 24, 2015

This expression is as rationalized as you will get it. No matter how you rearrange the expression, it will be an irrational number.

It seems that when people speak of rationalizing an expression of the form #a/b# then they mean rationalizing the denominator (bottom) - essentially moving any residual square roots into the numerator.

You can 'hide' it in a quadratic with integer coefficients as a root of that quadratic:

#(4x-sqrt(6)-3)(4x+sqrt(6)-3) = 16x^2-24x+3#

So #16x^2-24x+3=0# has roots

#x = (sqrt(6)-3)/4# and #(-sqrt(6)-3)/4#

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