Let . If and , then the quadratic equation having roots and is: [2025]

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Solution

Q.
Let $P_{n} = α^{n} + β^{n}, n \in N$ . If $P_{10} = 123, P_{9} = 76, P_{8} = 47$ and $P_{1} = 1$ , then the quadratic equation having roots $\frac{1}{α}$ and $\frac{1}{β}$ is: [2025]

1 $x^{2} + x - 1 = 0$

2 $x^{2} + x + 1 = 0$

3 $x^{2} - x + 1 = 0$

4 $x^{2} - x - 1 = 0$

Ans.
(1)

Since, $P_{10} = 123 \Rightarrow α^{10} + β^{10} = 123$

Since, $P_{9} = 76 \Rightarrow α^{9} + β^{9} = 76$

Since, $P_{8} = 47 \Rightarrow α^{8} + β^{8} = 47$

Since, $P_{1} = 1 \Rightarrow α + β = 1$

$∵ P_{1} \cdot P_{9} = P_{10} + α β P_{8}$

$\Rightarrow 1 \times 76 = 123 + α β (47)$

$\Rightarrow α β = - 1$

Also, $α + β = 1$

Clearly, $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{1}{- 1} = - 1, \frac{1}{α β} = - 1$

Thus, the required quadratic equation is $x^{2} + x - 1 = 0$ .

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