Question ID: #596
Let $P_{n}=\alpha^{n}+\beta^{n}, n \in \mathbb{N}$. 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:
- (1) $x^2-x+1=0$
- (2) $x^2+x-1=0$
- (3) $x^2-x-1=0$
- (4) $x^2+x+1=0$
Solution:
Let the quadratic equation with roots $\alpha, \beta$ be $x^2 – Sx + P = 0$, where $S = \alpha+\beta$ and $P = \alpha\beta$.
Given $P_1 = \alpha^1 + \beta^1 = 1$, so $S = 1$.
Using Newton’s Sum relation: $P_n – S P_{n-1} + P P_{n-2} = 0$.
Substitute $n=10$:
$$ P_{10} – S P_9 + P P_8 = 0 $$
Substitute the given values ($P_{10}=123, P_9=76, P_8=47, S=1$):
$$ 123 – 1(76) + P(47) = 0 $$
$$ 47 + 47P = 0 \Rightarrow P = -1 $$
So, $\alpha + \beta = 1$ and $\alpha\beta = -1$.
We need the equation for roots $\frac{1}{\alpha}, \frac{1}{\beta}$.
Sum of new roots:
$$ S’ = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{1}{-1} = -1 $$
Product of new roots:
$$ P’ = \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{-1} = -1 $$
The required quadratic equation is $x^2 – S’x + P’ = 0$:
$$ x^2 – (-1)x + (-1) = 0 $$
$$ x^2 + x – 1 = 0 $$
Ans. (2)
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