1. Let $P_n=\alpha^n+\beta^n.$ We are given: $P_{10}=123,\quad P_9=76,\quad P_8=47,\quad P_1=1.$

2. For numbers $\alpha,\beta$, the sequence $P_n$ satisfies the recurrence $P_n=(\alpha+\beta)P_{n-1}-\alpha\beta\,P_{n-2}.$ Let $s=\alpha+\beta,\quad p=\alpha\beta.$ Then $P_n=sP_{n-1}-pP_{n-2}.$

3. Use the given values for $n=10$: $P_{10}=sP_9-pP_8$ $123=76s-47p. \quad (1)$

4. Also, for $n=9$: $P_9=sP_8-pP_7$ but $P_7$ is unknown, so instead use the identity from lower terms.

   Since $P_0=\alpha^0+\beta^0=2,$ and $P_2=(\alpha+\beta)^2-2\alpha\beta=s^2-2p.$

   Also recurrence for $n=2$ gives $P_2=sP_1-pP_0=s-2p.$

   Equating both expressions for $P_2$: $s^2-2p=s-2p \Rightarrow s^2=s.$ Hence $s=0 \text{ or } 1.$

5. But given $P_1=\alpha+\beta=s=1.$ So, $\alpha+\beta=1.$

6. Now substitute $s=1$ into equation (1): $123=76-47p$ $47=-47p$ $p=-1.$ Therefore, $\alpha\beta=-1.$

7. The quadratic equation with roots $\alpha$ and $\beta$ is $x^2-(\alpha+\beta)x+\alpha\beta=0$ $x^2-x-1=0.$

8. We need the quadratic equation with roots $\dfrac1\alpha$ and $\dfrac1\beta$.

   Sum of these roots: $\frac1\alpha+\frac1\beta=\frac{\alpha+\beta}{\alpha\beta}=\frac{1}{-1}=-1.$

   Product of these roots: $\frac1\alpha\cdot\frac1\beta=\frac1{\alpha\beta}=-1.$

   Hence the required quadratic is $x^2-(\text{sum})x+(\text{product})=0$ $x^2-(-1)x+(-1)=0$ $x^2+x-1=0.$

9. Checking options:

   • A: $x^2+x-1=0$ ✓
   • B: $x^2-x+1=0$ ✗
   • C: $x^2+x+1=0$ ✗
   • D: $x^2-x-1=0$ ✗

Therefore, the correct option is A.