MathRevolution wrote:

Which of the following is equal to \(1/(√3−√2)^2\)?

A) 1

B) 5

C) √6

D) 5 - √6

E) 5 + 2√6

I. Simplify and rationalizeSimplify:

\(\frac{1}{(√3−√2)^2}\)

\(\frac{1}{(√3−√2)(√3−√2)}\)

\(\frac{1}{3-2√6+2}=\frac{1}{5-2√6}\)

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 onceThe 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

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