If x = 3 - 2sqrt{2},then sqrt{x} + frac{1}{sq | vedclass

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Question

(D) Given $x = 3 - 2\sqrt{2}$.We can write $x$ as a perfect square: $x = 2 + 1 - 2\sqrt{2} = (\sqrt{2})^2 + (1)^2 - 2(\sqrt{2})(1) = (\sqrt{2} - 1)^2$

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$2\sqrt{2}$

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(D) Given $x = 3 - 2\sqrt{2}$.We can write $x$ as a perfect square: $x = 2 + 1 - 2\sqrt{2} = (\sqrt{2})^2 + (1)^2 - 2(\sqrt{2})(1) = (\sqrt{2} - 1)^2$.Therefore,$\sqrt{x} = \sqrt{(\sqrt{2} - 1)^2} = \sqrt{2} - 1$.Now,find $\frac{1}{\sqrt{x}}$:$\frac{1}{\sqrt{x}} = \frac{1}{\sqrt{2} - 1}$.Rationalizing the denominator:$\frac{1}{\sqrt{2} - 1} 	imes \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1$.Finally,calculate $\sqrt{x} + \frac{1}{\sqrt{x}}$:$(\sqrt{2} - 1) + (\sqrt{2} + 1) = 2\sqrt{2}$.

If $x = 3 - 2\sqrt{2}$,then $\sqrt{x} + \frac{1}{\sqrt{x}}$ is equal to

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