If a = 7 - 4 sqrt{3},the value of a^{frac{1}{ | vedclass

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(B) Given $a = 7 - 4 \sqrt{3}$.First,find $\frac{1}{a}$ by rationalizing the denominator:$\frac{1}{a} = \frac{1}{7 - 4 \sqrt{3}} 	imes \frac{7 + 4 \s

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$4$

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(B) Given $a = 7 - 4 \sqrt{3}$.First,find $\frac{1}{a}$ by rationalizing the denominator:$\frac{1}{a} = \frac{1}{7 - 4 \sqrt{3}} 	imes \frac{7 + 4 \sqrt{3}}{7 + 4 \sqrt{3}} = \frac{7 + 4 \sqrt{3}}{49 - 16(3)} = \frac{7 + 4 \sqrt{3}}{49 - 48} = 7 + 4 \sqrt{3}$.We need to find $a^{\frac{1}{2}} + a^{-\frac{1}{2}}$,which is $\sqrt{a} + \frac{1}{\sqrt{a}}$.Let $x = \sqrt{a} + \frac{1}{\sqrt{a}}$.Then $x^2 = (\sqrt{a} + \frac{1}{\sqrt{a}})^2 = a + \frac{1}{a} + 2$.Substitute the values of $a$ and $\frac{1}{a}$:$x^2 = (7 - 4 \sqrt{3}) + (7 + 4 \sqrt{3}) + 2 = 14 + 2 = 16$.Since $a > 0$,$\sqrt{a} + \frac{1}{\sqrt{a}}$ must be positive,so $x = \sqrt{16} = 4$.

If $a = 7 - 4 \sqrt{3}$,the value of $a^{\frac{1}{2}} + a^{-\frac{1}{2}}$ is

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