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

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

Answer: 4

The value of

\sqrt{x} + \frac{1}{\sqrt{x}}

is 4.

Expressing

7 - 4\sqrt{3}

(2 - \sqrt{3})^2

allows taking the square root to get

\sqrt{x} = 2 - \sqrt{3}

. Rationalizing its reciprocal gives

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

. Adding these two values cancels the irrational component

\sqrt{3}

, leaving

2 + 2 = 4

Simplify the nested surd

\sqrt{x} = \sqrt{7 - 4\sqrt{3}}

\sqrt{x} = 2 - \sqrt{3}

Rewrite

7 - 4\sqrt{3}

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

and take the principal square root.

Calculate the reciprocal

\frac{1}{\sqrt{x}}

by rationalizing the denominator

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

Multiply numerator and denominator of

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

by its conjugate

(2 + \sqrt{3})

Add

\sqrt{x}

and

\frac{1}{\sqrt{x}}

Sum

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

so that the radical terms cancel out.

Simplification of surds of the form

\sqrt{a \pm \sqrt{b}}

and rationalization using conjugates

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