Let y=2+2,z=2−2 and x=y+z, then x is

Given y=2+2,z=2−2∴x=y+z=2+2+2−2x2=(2+2)+(2−2)+2(2+2)(2−2)x2=4+22x2=2(2+2)x=2.2+2

Let y=2+2,z=2−2 and x=y+z, then x is

Given y=2+2,z=2−2∴x=y+z=2+2+2−2x2=(2+2)+(2−2)+2(2+2)(2−2)x2=4+22x2=2(2+2)x=2.2+2

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Let $y = \sqrt{2 + \sqrt{2}}, z = \sqrt{2 - \sqrt{2}}$ and $x = y + z$ , then $x$ is

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$\sqrt{2 + \sqrt{2}}$

$\sqrt{2} . \sqrt{2 + \sqrt{2}}$

$\sqrt{2} . \sqrt{2 - \sqrt{2}}$

$\sqrt{\sqrt{2} - \sqrt{2 + \sqrt{2}}}$

answer is C.

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Detailed Solution

Given $y = \sqrt{2 + \sqrt{2}}, z = \sqrt{2 - \sqrt{2}}$

$∴ x = y + z = \sqrt{2 + \sqrt{2}} + \sqrt{2 - \sqrt{2}}$

$x^{2} = (2 + \sqrt{2}) + (2 - \sqrt{2}) + 2 \sqrt{(2 + \sqrt{2}) (2 - \sqrt{2})}$

$x^{2} = 4 + 2 \sqrt{2}$

$x^{2} = 2 (2 + \sqrt{2})$

$x = \sqrt{2} . \sqrt{2 + \sqrt{2}}$

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