If x = \sqrt{7 + 4\sqrt{3}}, what is the value of x - \frac{1}{x}? — GMAT

If $x = \sqrt{7 + 4\sqrt{3}}$ , what is the value of $x - \frac{1}{x}$ ?

2\sqrt{3}

The expression under the outer radical,

7 + 4\sqrt{3}

, can be rewritten as

(2 + \sqrt{3})^2

because

2^2 + 2(2)(\sqrt{3}) + (\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}

. Taking the square root gives

x = 2 + \sqrt{3}

. The reciprocal

\frac{1}{x}

\frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}

. Thus,

x - \frac{1}{x} = (2 + \sqrt{3}) - (2 - \sqrt{3}) = 2\sqrt{3}

Un-nest the radical

\sqrt{7 + 4\sqrt{3}}

by expressing

7 + 4\sqrt{3}

as a perfect square

(a + b\sqrt{3})^2

(a + b\sqrt{3})^2 = a^2 + 3b^2 + 2ab\sqrt{3} = 7 + 4\sqrt{3}

, which yields

a = 2

and

b = 1

, so

x = 2 + \sqrt{3}

Recognizing nested radicals in the form

\sqrt{A + B\sqrt{C}}

allows simplification into a binomial radical.

Find the reciprocal

\frac{1}{x}

by rationalizing the denominator.

\frac{1}{2 + \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}.

Multiplying the numerator and denominator by the conjugate clears the radical from the denominator.

Calculate the difference

x - \frac{1}{x}

(2 + \sqrt{3}) - (2 - \sqrt{3}) = 2 + \sqrt{3} - 2 + \sqrt{3} = 2\sqrt{3}.

Subtracting the reciprocal isolates the irrational component.

Simplification of nested radicals and rationalizing denominators using algebraic conjugates.

Estimated Time:1m 30s

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