_8=47,\quad P_1=1.$$

We want the quadratic equation whose roots are $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$.

For numbers $\alpha,\beta$, the sequence $P_n=\alpha^n+\beta^n$ satisfies

$$P_n=(\alpha+\beta)P_{n-1}-(\alpha\beta)P_{n-2}.$$

Now use the given values:

$$P_{10}=(\alpha+\beta)P_9-(\alpha\beta)P_8.$$

So,

$$123=(\alpha+\beta)\cdot 76-(\alpha\beta)\cdot 47.$$

Also, since $P_1=\alpha+\beta=1$, we get

$$123=76-47\alpha\beta.$$

Thus,

$$47=-47\alpha\beta,$$

$$\alpha\beta=-1.$$

Now the roots needed are $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$.

Their sum is

$$\frac{1}{\alpha}+\frac{1}{\beta} =\frac{\alpha+\beta}{\alpha\beta} =\frac{1}{-1}=-1.$$

Their product is

$$\frac{1}{\alpha\beta}=\frac{1}{-1}=-1.$$

A quadratic equation with roots $r_1,r_2$ is

$$x^2-(r_1+r_2)x+r_1r_2=0.$$

Here,

$$x^2-(-1)x+(-1)=0,$$

which gives

$$x^2+x-1=0.$$

So the correct option is:

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

Option A.

19:T71c,

दिया गया है

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

और

$$P_{10}=123,\quad P_9=76,\quad P_8=47,\quad P_1=1.$$

हमें उस द्विघात समीकरण की आवश्यकता है जिसके मूल $\dfrac{1}{\alpha}$ और $\dfrac{1}{\beta}$ हैं।

संख्याओं $\alpha,\beta$ के लिए, अनुक्रम $P_n=\alpha^n+\beta^n$ इस पुनरावृत्ति संबंध का पालन करता है

$$P_n=(\alpha+\beta)P_{n-1}-(\alpha\beta)P_{n-2}.$$

अब दिए गए मानों का उपयोग करें:

$$P_{10}=(\alpha+\beta)P_9-(\alpha\beta)P_8.$$

तो,

$$123=(\alpha+\beta)\cdot 76-(\alpha\beta)\cdot 47.$$

चूंकि $P_1=\alpha+\beta=1$, अतः

$$123=76-47\alpha\beta.$$

अर्थात्,

$$47=

Quadratic Equation And Inequalities

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