Let alpha and beta are roots | vedclass

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Question

$\alpha  + \beta  = \frac{3}{5}$

${\alpha \beta}  {=-\frac{1}{5}}$

$\left[ {(\alpha  + \beta )x - \left( {\frac{{{\alpha ^2}

Vedclass · Accepted Answer

none of these

Vedclass · Answer

$\alpha  + \beta  = \frac{3}{5}$

${\alpha \beta}  {=-\frac{1}{5}}$

$\left[ {(\alpha  + \beta )x - \left( {\frac{{{\alpha ^2} + {\beta ^2}}}{2}} \right){x^2} + \left( {\frac{{{\alpha ^3} + {\beta ^3}}}{3}} \right){x^3} \cdots  \cdots  \cdots } \right]$

$=\left(\alpha \mathrm{x}-\frac{(\alpha \mathrm{x})^{2}}{2}+\frac{(\alpha \mathrm{x})^{3}}{3} \ldots \ldots \ldots\right)+\left(\beta \mathrm{x}-\frac{(\beta \mathrm{x})^{2}}{2}+\frac{(\beta \mathrm{x})^{3}}{3} \ldots \ldots\right)$

$=\ln (1-\alpha x)+ln(1-\beta x)$

$=\ln ((1-\alpha x)(1-\beta x))$

$ = l{\rm{n}}\left( {1 - {\rm{x}}(\alpha  + \beta ) + \alpha \beta {{\rm{x}}^2}} \right)$

$=\ln \left(1-x \frac{3}{5}-\frac{1}{5} x^{2}\right)$

Let $\alpha $ and $\beta $ are roots of $5{x^2} - 3x - 1 = 0$ , then $\left[ {\left( {\alpha + \beta } \right)x - \left( {\frac{{{\alpha ^2} + {\beta ^2}}}{2}} \right){x^2} + \left( {\frac{{{\alpha ^3} + {\beta ^3}}}{3}} \right){x^3} -......} \right]$ is

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