If x=√6+2 and y=√6-2, then what is the value of (x/y+y/x)^2-3?

Question

If x=√6+2 and y=√6-2, then what is the value of (x/y+y/x)^2-3?

1. A. 22
2. B. 35
3. C. 42
4. D. 97 (Correct answer)

Correct Answer

Option D — 97

Detailed Solution & Explanation

The correct answer is 97.

Key Points

• $x^2+y^2=(\sqrt6+2)^2+(\sqrt6-2)^2=(10+4\sqrt6)+(10-4\sqrt6)=20$ and $xy=(\sqrt6)^2-2^2=6-4=2$.
• $\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{x^2+y^2}{xy}=\dfrac{20}{2}=10$.
• Required value $=10^2-3=97$.

Additional Information

• The expression x/y + y/x is always (x² + y²)/xy, so compute those two quantities rather than the fractions themselves.
• For conjugate surds the work collapses: (a+b)² + (a−b)² = 2(a² + b²) and (a+b)(a−b) = a² − b², both of which clear the radicals.
• Here x² + y² = 20 and xy = 2 — neither contains a surd, which is the signal that the intended route is correct.