Question

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

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

Answer 1

Answer: (C)

$4-\sqrt{5}$

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