Question

Let \( x = -4\sqrt{2} + \sqrt{17(-\sqrt{2})^2 + 2} \). If \( \frac{1}{x} = a + b\sqrt{2} \), then what is the value of \( (a - b) \)?

• A. \( 0 \)
• B. \( \frac{1}{2} \)
• C. \( 1 \)
• D. \( \frac{3}{2} \)

Hint:

When irrational numbers are involved, rationalize the denominator and simplify to match standard forms.

Answer 1

The Correct Option is B

We simplify: \( x = -4\sqrt{2} + \sqrt{17 \cdot 2 + 2} = -4\sqrt{2} + \sqrt{34 + 2} = -4\sqrt{2} + \sqrt{36} = -4\sqrt{2} + 6 \) So, \( \frac{1}{x} = \frac{1}{6 - 4\sqrt{2}} \). Multiply numerator and denominator by conjugate: \( \frac{1}{6 - 4\sqrt{2}} \cdot \frac{6 + 4\sqrt{2}}{6 + 4\sqrt{2}} = \frac{6 + 4\sqrt{2}}{(6)^2 - (4\sqrt{2})^2} = \frac{6 + 4\sqrt{2}}{36 - 32} = \frac{6 + 4\sqrt{2}}{4} = \frac{3}{2} + \sqrt{2} \) Thus, \( a = \frac{3}{2}, b = 1 \Rightarrow a - b = \frac{3}{2} - 1 = \frac{1}{2} \)