Discussion

Question:

If x = 8 - $\sqrt{32}$ and y = 2 + √2, then ${(x + \frac{1}{y})}^{2}$ is given by:

Explanation:

x = 8 − √32 = 8 − 4√2 = 4(2 − √2)

$\frac{1}{y}$ = $\frac{1}{2 + \sqrt{2}}$

On rationalising, we get

$\frac{1}{y}$ = $\frac{1}{2 + \sqrt{2}} \times \frac{2 - \sqrt{2}}{2 - \sqrt{2}}$ = $\frac{2 - \sqrt{2}}{2}$ = $\frac{x}{8}$

∴ x + 1/y = x + x/8 = 9x/8

⇒ ${(x + \frac{1}{y})}^{2}$ = ${(x + \frac{x}{8})}^{2}$ = ${(\frac{9 x}{8})}^{2}$ = $\frac{81 x^{2}}{64}$

Hence option (d).

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