Let P_n= ^n+ ^n,n N.

Question

Let ${P}_{n}={\alpha }^{n}+{\beta }^{n},n\in N$. If ${P}_{10}=123,{P}_{9}=76,{P}_{8}=47$ and ${P}_{1}=1$, then the quadratic equation having roots $\frac{1}{\alpha }$ and $\frac{1}{\beta }$ is :

Accepted Answer

Given:

$ P_{10} = 123 $

$ P_9 = 76 $

$ P_8 = 47 $

$ P_1 = 1 $

We know that:

$ P_n = \alpha^n + \beta^n $

According to Newton's identities, we have the relation:

$ P_{10} = P_9 + P_8 $

This implies:

$ P_{10} - P_9 - P_8 = 0 $

From $ P_1 = 1 $, it follows that:

$ \alpha + \beta = 1 $

$ \alpha \beta = 1 $

Now, we are tasked with finding the quadratic equation whose roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$. For such a quadratic equation:

Using the relationship between roots and coefficients, the equation with roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is:

$ x^2 - \left(\frac{1}{\alpha} + \frac{1}{\beta}\right)x + \frac{1}{\alpha \beta} = 0 $

Substitute the known sum and product of $\alpha$ and $\beta$:

$ x^2 - \left(\frac{\alpha + \beta}{\alpha \beta}\right)x + \frac{1}{\alpha \beta} = 0 $

Given that $\alpha + \beta = 1$ and $\alpha \beta = 1$, we can simplify:

$ x^2 - \left(\frac{1}{1}\right)x + \frac{1}{1} = 0 $

This simplifies to:

$ x^2 + x - 1 = 0 $

Thus, the quadratic equation having roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is:

$ x^2 + x - 1 = 0 $

Solutions · 1