Let $\mathrm{P}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}, \mathrm{n} \in \mathrm{N}$. If $\mathrm{P}_{10}=123, \mathrm{P}_9=76, \mathrm{P}_8=47$ and $\mathrm{P}_1=1$, then the quadratic equation having roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is :

<p>Given:</p> <p><p>$ P_{10} = 123 $</p></p> <p><p>$ P_9 = 76 $</p></p> <p><p>$ P_8 = 47 $</p></p> <p><p>$ P_1 = 1 $</p></p> <p>We know that:</p> <p>$ P_n = \alpha^n + \beta^n $</p> <p>According to Newton’s identities, we have the relation:</p> <p>$ P_{10} = P_9 + P_8 $</p> <p>This implies:</p> <p>$ P_{10} - P_9 - P_8 = 0 $</p> <p>From $ P_1 = 1 $, it follows that:</p> <p>$ \alpha + \beta = 1 $</p> <p>$ \alpha \beta = 1 $</p> <p>Now, we are tasked with finding the quadratic equation whose roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$. For such a quadratic equation:</p> <p>Using the relationship between roots and coefficients, the equation with roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is:</p> <p>$ x^2 - \left(\frac{1}{\alpha} + \frac{1}{\beta}\right)x + \frac{1}{\alpha \beta} = 0 $</p> <p>Substitute the known sum and product of $\alpha$ and $\beta$:</p> <p>$ x^2 - \left(\frac{\alpha + \beta}{\alpha \beta}\right)x + \frac{1}{\alpha \beta} = 0 $</p> <p>Given that $\alpha + \beta = 1$ and $\alpha \beta = 1$, we can simplify:</p> <p>$ x^2 - \left(\frac{1}{1}\right)x + \frac{1}{1} = 0 $</p> <p>This simplifies to:</p> <p>$ x^2 + x - 1 = 0 $</p> <p>Thus, the quadratic equation having roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is:</p> <p>$ x^2 + x - 1 = 0 $</p>