If A.M. of the roots of a quadratic equation | vedclass

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Question

(A) Let the roots be $\alpha$ and $\beta$.The $A.M.$ of the roots is $\frac{\alpha + \beta}{2} = \frac{8}{5}$,which implies $\alpha + \beta = \frac{16

Vedclass · Accepted Answer

$5x^2 - 16x + 7 = 0$

Vedclass · Answer

(A) Let the roots be $\alpha$ and $\beta$.The $A.M.$ of the roots is $\frac{\alpha + \beta}{2} = \frac{8}{5}$,which implies $\alpha + \beta = \frac{16}{5}$ $(i)$.The $A.M.$ of their reciprocals is $\frac{\frac{1}{\alpha} + \frac{1}{\beta}}{2} = \frac{8}{7}$.This simplifies to $\frac{\alpha + \beta}{2\alpha\beta} = \frac{8}{7}$.Substituting $\alpha + \beta = \frac{16}{5}$ into the equation: $\frac{16/5}{2\alpha\beta} = \frac{8}{7} \Rightarrow \frac{8}{5\alpha\beta} = \frac{8}{7}$.Thus,$\alpha\beta = \frac{7}{5}$ $(ii)$.The quadratic equation is given by $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Substituting the values: $x^2 - (\frac{16}{5})x + \frac{7}{5} = 0$.Multiplying by $5$,we get $5x^2 - 16x + 7 = 0$.

If $A.M.$ of the roots of a quadratic equation is $8/5$ and $A.M.$ of their reciprocals is $8/7$,then the equation is

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