If the harmonic mean of the roots of the equa | vedclass

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(C) Given the quadratic equation: $\sqrt{2}x^2 - bx + (8 - 2\sqrt{5}) = 0$. Let the roots of the equation be $\alpha$ and $\beta$. From the properties

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$4 - \sqrt{5}$

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(C) Given the quadratic equation: $\sqrt{2}x^2 - bx + (8 - 2\sqrt{5}) = 0$. Let the roots of the equation be $\alpha$ and $\beta$. From the properties of quadratic equations,the sum of the roots is $\alpha + \beta = \frac{-(-b)}{\sqrt{2}} = \frac{b}{\sqrt{2}}$. The product of the roots is $\alpha\beta = \frac{8 - 2\sqrt{5}}{\sqrt{2}}$. The harmonic mean $(HM)$ of two roots is given by $HM = \frac{2\alpha\beta}{\alpha + \beta}$. Given $HM = 4$,we have: $\frac{2\alpha\beta}{\alpha + \beta} = 4$ Substituting the values of $\alpha + \beta$ and $\alpha\beta$: $\frac{2 \left( \frac{8 - 2\sqrt{5}}{\sqrt{2}} \right)}{\frac{b}{\sqrt{2}}} = 4$ $\frac{2(8 - 2\sqrt{5})}{b} = 4$ $2(8 - 2\sqrt{5}) = 4b$ $8 - 2\sqrt{5} = 2b$ $b = 4 - \sqrt{5}$.

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

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