QQuestion
If x = √3 + √2 and y = 1/x then find (x + \frac{1}{\text{x}})² + (\frac{1}{\text{y}} – y)²
✓Answer
x = √3 + √2
⇒ \frac{1}{\text{x}} = \frac{1}{\text{√3 + √2}}
= \frac{1}{\text{√3 + √2}} × \frac{√3 - √2}{\text{√3 - √2}}
= \frac{√3 - √2}{\text{(√3)² - (√2)²}}
\frac{1}{\text{x}} = √3 – √2
Also, y = \frac{1}{\text{x}} = √3 – √2 and \frac{1}{\text{y}} = x = √3 + √2
Now, (x + \frac{1}{\text{x}})² + (\frac{1}{\text{y}} – y)² = {(√3 + √2) + (√3 – √2)}² + {(√3 + √2) – (√3 – √2)}²
= (2√3)² + (2√2)²
= 12 + 8
= 20
∴ (x + \frac{1}{\text{x}})² + (\frac{1}{\text{y}} – y)² = 20
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