Harmonic mean of the roots of the equation (5 | vedclass

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(C) For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$,the harmonic mean $H$ is given by the formula $H = \frac{2\alpha\beta

Vedclass · Accepted Answer

$4$

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(C) For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$,the harmonic mean $H$ is given by the formula $H = \frac{2\alpha\beta}{\alpha + \beta}$.From the given equation $(5 + \sqrt{2})x^2 - (4 + \sqrt{5})x + 8 + 2\sqrt{5} = 0$,we identify the coefficients as $a = 5 + \sqrt{2}$,$b = -(4 + \sqrt{5})$,and $c = 8 + 2\sqrt{5}$.Using Vieta's formulas,the product of the roots $\alpha\beta = \frac{c}{a} = \frac{8 + 2\sqrt{5}}{5 + \sqrt{2}}$ and the sum of the roots $\alpha + \beta = -\frac{b}{a} = \frac{4 + \sqrt{5}}{5 + \sqrt{2}}$.Substituting these into the formula for $H$:$H = \frac{2(\frac{8 + 2\sqrt{5}}{5 + \sqrt{2}})}{\frac{4 + \sqrt{5}}{5 + \sqrt{2}}} = \frac{2(8 + 2\sqrt{5})}{4 + \sqrt{5}}$.Factoring out $2$ from the numerator: $H = \frac{2 \cdot 2(4 + \sqrt{5})}{4 + \sqrt{5}} = 4$.

Harmonic mean of the roots of the equation $(5 + \sqrt{2})x^2 - (4 + \sqrt{5})x + 8 + 2\sqrt{5} = 0$ is:

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