If the harmonic mean between the roots of (5+ | vedclass

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(D) Given equation is $(5+\sqrt{2}) x^2-b x+(8+2 \sqrt{5})=0$.Let $\alpha$ and $\beta$ be the roots of this equation.From the relation between roots a

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$4+\sqrt{5}$

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(D) Given equation is $(5+\sqrt{2}) x^2-b x+(8+2 \sqrt{5})=0$.Let $\alpha$ and $\beta$ be the roots of this equation.From the relation between roots and coefficients:$\alpha+\beta = \frac{b}{5+\sqrt{2}}$$\alpha \beta = \frac{8+2 \sqrt{5}}{5+\sqrt{2}}$The harmonic mean $(HM)$ between the roots is given by $HM = \frac{2 \alpha \beta}{\alpha+\beta}$.Given $HM = 4$,we have:$\frac{2 \alpha \beta}{\alpha+\beta} = 4$Substituting the values of $\alpha+\beta$ and $\alpha \beta$:$\frac{2 	imes \frac{8+2 \sqrt{5}}{5+\sqrt{2}}}{\frac{b}{5+\sqrt{2}}} = 4$$\frac{2(8+2 \sqrt{5})}{b} = 4$$\frac{8+2 \sqrt{5}}{b} = 2$$b = \frac{8+2 \sqrt{5}}{2} = 4+\sqrt{5}$.

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

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