Question

Let $p,q$ be integers and let $\alpha ,\beta$ be the roots of the equation $x^2–2x+3=0$ where $\alpha \ne \beta .$ For $n=0,1,2,...,$ let $a_n=p{\alpha }^n+q{\beta }^n$ and $a_3=-10,$ then $p^q+q^p=$ ___

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is A

2

Since, $\alpha$ and $\beta$ are the roots of the equation, $x^2-2x+3=0$,

so, $\alpha +\beta =2$ and $\alpha \beta =3$

Also, ${\alpha }^2-2\alpha +3=0$ and ${\beta }^2-2\beta +3=0$

$\Rightarrow {\alpha }^2=2\alpha -3$ and ${\beta }^2=2\beta -3$

Given, $a_n=p{\alpha }^n+q{\beta }^n$

So, $a_0=p+q$

$a_1=p\alpha +q\beta$

$a_2=p{\alpha }^2+q{\beta }^2$

$=p(2\alpha -3)+q(2\beta -3)$

$=2(p\alpha +q\beta )-3(p+q)$

$=2a_1-3a_0$

Similarly, $a_3=2a_2-3a_1$

$=2(2a_1-3a_0)-3a_1$

$=a_1-6a_0$

$\Rightarrow -10=p\alpha +q\beta -6(p+q)$

$-10=p\alpha +q(2-\alpha )-6(p+q)$

$-10=\alpha (p-q)+(-6p-4q)$

$\Rightarrow p=q;-10q=-10$

$\Rightarrow q=p=1$

Thus, $p^q+q^p=1+1=2$