Question

If the harmonic mean of the roots of$\sqrt{2}{x}^2-bx+(8-2\sqrt{5})=0$ is 4, the the value of b=

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

Answer 1

The correct option is C $4-\sqrt{5}$

$Let\alpha ,\beta betheroots$
$\Rightarrow \frac{2\alpha \beta }{\alpha +\beta }=4\Rightarrow \frac{\frac{2(8-2\sqrt{5})}{\sqrt{2}}}{\frac{b}{\sqrt{2}}}=4$
$\Rightarrow \frac{2(8-2\sqrt{5})}{4}=b$
$\therefore b=4-\sqrt{5}$