Which of the following is equal to 1/[(3^(1/2)-2^(1/2))^1/2]

I. Simplify and rationalize
Simplify:
\(\frac{1}{(√3−√2)^2}\)
\(\frac{1}{(√3−√2)(√3−√2)}\)
\(\frac{1}{3-2√6+2}=\frac{1}{5-2√6}\)

Rationalize with conjugate: \(5+2√6\)
\((\frac{1}{5-2√6}*\frac{5+2√6}{5+2√6})=\frac{5+2√6}{25-(4*6)}\)

\(\frac{5+2√6}{25-24}=\frac{5+2√6}{1}=5+2√6\)

Answer E

Rationalize all at once
The denominator has two "copies" of \((a-b)\), one of the factored terms of a difference of perfect square \((a+b)(a-b)=a^2-b^2\).

If we multiply by two copies of \((a+b)\) -- the conjugate of the denominator, squared -- we will create two differences of perfect squares that in this instance are integers.

\(\frac{1}{(√3−√2)^2}=\frac{1}{(√3−√2)(√3−√2)}\)

\(\frac{1}{(√3−√2)(√3−√2)}*\frac{(√3+√2)^2}{(√3+√2)^2}\)

\(\frac{1}{(√3−√2)(√3−√2)}*\frac{(√3+√2)(√3+√2)}{(√3+√2)(√3+√2)}\)

\(\frac{(√3+√2)(√3+√2)}{(√3−√2)(√3+√2)(√3−√2)(√3+√2)}\)

\(\frac{3+2√6+2}{(3-2)(3-2)}=\frac{5+2√6}{(1*1)}=5+2√6\)

Answer E

Can you please let me know why did we multiply both nominator and denominator by \(5+2\sqrt{6}\)?

Thanks!

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