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

$\begin{array}{l}Let\alpha ,\beta arerootsof\sqrt{2}{x}^2-bx+(8+2\sqrt{5})=0\\ GivenH.Mof\alpha ,\beta =4\\ \Rightarrow \frac{2\alpha \beta }{\alpha +\beta }=4\\ \Rightarrow \frac{2(8-2\sqrt{5})/\sqrt{2}}{\left(\frac{b}{\sqrt{2}}\right)}=4\\ \Rightarrow 2(8-2\sqrt{5})=4\left(\frac{b}{c}\right)\\ \Rightarrow b=\frac{(8-2\sqrt{5})}{2}\\ \Rightarrow 4-\sqrt{5}\end{array}$