Question 345955
I assume by "root" you mean square root. Please be specific in the future because there are also cube roots, 4th roots, 5th roots, etc.<br>
{{{(sqrt(3)-2)/(-5sqrt(3)+8)}}}
To rationalize a binomial (two-term) denominator with square roots, we will take advantage of the pattern: {{{(a+b)(a-b) = a^2 - b^2}}}. This shows us how to multiply a two term expression like a+b, a-b, or your denominator and turn it into an expression of perfect squares!. You denominator has a "+" between the two terms so it will play the role of a+b with "a" being {{{-5sqrt(3)}}} and "b" being 8. So we need to multiply by the corresponding a-b: {{{-5sqrt(3)-8}}}. And since we are multiplying the denominator by this, we must also multiply the numerator:
{{{((sqrt(3)-2)/(-5sqrt(3)+8))((-5sqrt(3)-8)/(-5sqrt(3)-8))}}}
When we multiply this out, the denominator is easy because we know from the pattern how it works out: {{{a^2 - b^2}}}} or, in this case: {{{(-5sqrt(3))^2 - (8)^2}}}. In the numerator we will just use FOIL (or whatever method you have learned to multiply binomials).
{{{(sqrt(3)*(-5sqrt(3)) -sqrt(3)*8 + (-2)(-5sqrt(3)) + (-2)*8)/((-5sqrt(3))^2 - (8)^2)}}}
{{{(-5*3 -8sqrt(3) +10sqrt(3) -16)/(25*3 - 64)}}}
{{{(-15  -8sqrt(3) +10sqrt(3) -16)/(75 - 64)}}}
{{{(-31 +2sqrt(3))/11}}}
And we have a simplified fraction with a rational denominator.