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

Vedclass · Accepted Answer

$4 - \sqrt{5}$

Vedclass · Answer

(C) Let the roots of the quadratic equation $\sqrt{2}x^2 - bx + (8 - 2\sqrt{5}) = 0$ be $\alpha$ and $\beta$.From the properties of quadratic equations,the sum of roots $\alpha + \beta = \frac{b}{\sqrt{2}}$ and the product of roots $\alpha\beta = \frac{8 - 2\sqrt{5}}{\sqrt{2}} = 4\sqrt{2} - \sqrt{10}$.The harmonic mean $(HM)$ of two roots is given by $HM = \frac{2\alpha\beta}{\alpha + \beta}$.Given $HM = 4$,we have $4 = \frac{2(4\sqrt{2} - \sqrt{10})}{\frac{b}{\sqrt{2}}}$.$4 = \frac{2(4\sqrt{2} - \sqrt{10}) \cdot \sqrt{2}}{b}$.$4 = \frac{2(8 - \sqrt{20})}{b} = \frac{2(8 - 2\sqrt{5})}{b} = \frac{16 - 4\sqrt{5}}{b}$.$4b = 16 - 4\sqrt{5}$.Dividing by $4$,we get $b = 4 - \sqrt{5}$.

List-$I$	List-$II$
$(i) \alpha = \beta$	$(A) (ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$
$(ii) \alpha = 2\beta$	$(B) 2b^2 = 9ac$
$(iii) \alpha = 3\beta$	$(C) b^2 = 6ac$
$(iv) \alpha = \beta^2$	$(D) 3b^2 = 16ac$
	$(E) b^2 = 4ac$
	$(F) (ac^2)^{1/3} + (a^2c)^{1/3} = b$

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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