Theory of Equations 1 Question 55
56. If $\alpha$ and $\beta$ are the roots of the equation $x^{2}+p x+1=0 ; \gamma, \delta$ are the roots of $x^{2}+q x+1=0$, then $q^{2}-p^{2}=(\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta)$
$(1978,2 M)$
Passage Type Questions
Let $p, q$ be integers and let $\alpha, \beta$ be the roots of the equation, $x^{2}-x-1=0$ where $\alpha \neq \beta$. For $n=0,1,2, \ldots \ldots$, let $a _n=p \alpha^{n}+q \beta^{n}$.
FACT : If $a$ and $b$ are rational numbers and $a+b \sqrt{5}=0$, then $a=0=b$.
(2017 Adv.)
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Solution:
- Since, $\alpha+\beta=-p, \alpha \beta=1$ and $\gamma+\delta=-q, \gamma \delta=1$
Now, $(\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta)$
$$ \begin{aligned} & ={\alpha \beta-\gamma(\alpha+\beta)+\gamma^{2} }{\alpha \beta+\delta(\alpha+\beta)+\delta^{2} } \\ & ={1-\gamma(-p)+\gamma^{2} }{1+\delta(-p)+\delta^{2} } \end{aligned} $$
$$ \begin{gathered} =\left(1+\gamma^{2}+\gamma p\right)\left(1-\delta p+\delta^{2}\right)=(-q \gamma+\gamma p)(-\delta p-\delta q) \\ {\left[\because \gamma^{2}+q \gamma+1=0 \text { and } \delta^{2}+q \delta+1=0\right]} \\ =\left(q^{2}-p^{2}\right)(\gamma \delta)=q^{2}-p^{2} \end{gathered} $$
