Question

If the harmonic mean of the roots of the equation \[ \sqrt{2} x^2 - b x + \left( 8 - 2\sqrt{5} \right) = 0 \] is 4, then the value of \( b \) is:

• A. \(3\)
• B. \(2\)
• C. \(4 - \sqrt{5}\)
• D. \(4 + \sqrt{5}\)

Hint:

Harmonic mean of roots: \( HM = \frac{2 \alpha \beta}{\alpha + \beta} \).

Answer 1

The Correct Option is C

Step 1: Use quadratic formula terms
For quadratic \( a x^2 + b x + c = 0 \), roots are \( \alpha \) and \( \beta \). Step 2: Sum and product of roots
\[ \alpha + \beta = \frac{-b}{a}, \quad \alpha \beta = \frac{c}{a} \] Here, \( a = \sqrt{2} \), \( b = -b \), \( c = 8 - 2\sqrt{5} \). Sum: \[ \alpha + \beta = \frac{b}{\sqrt{2}} \] Product: \[ \alpha \beta = \frac{8 - 2\sqrt{5}}{\sqrt{2}} \] Step 3: Harmonic mean (HM)
\[ HM = \frac{2}{\frac{1}{\alpha} + \frac{1}{\beta}} = \frac{2 \alpha \beta}{\alpha + \beta} = 4 \] Substitute: \[ \frac{2 \times \frac{8 - 2\sqrt{5}}{\sqrt{2}}}{\frac{b}{\sqrt{2}}} = 4 \implies \frac{2(8 - 2\sqrt{5})}{b} = 4 \] Step 4: Solve for \( b \)
\[ \frac{2(8 - 2\sqrt{5})}{b} = 4 \implies 2(8 - 2\sqrt{5}) = 4b \implies b = \frac{2(8 - 2\sqrt{5})}{4} = \frac{8 - 2\sqrt{5}}{2} = 4 - \sqrt{5} \]