Theory of Equations 1 Question 53
54. If $\alpha$ and $\beta$ are the roots of $x^{2}+p x+q=0$ and $\gamma, \delta$ are the roots of $x^{2}+r x+s=0$, then evaluate $(\alpha-\gamma)(\beta-\gamma)$ $(\alpha-\delta)(\beta-\delta)$ in terms of $p, q, r$ and $s$.
$(1979,2 M)$
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Answer:
Correct Answer: 54. $((q-s)^{2}-r q p-r s p+s p^{2}+q r^{2})$
Solution:
- Since, $\alpha, \beta$ are the roots of $x^{2}+p x+q=0$
and $\gamma, \delta$ are the roots of $x^{2}+r x+s=0$
$ \begin{array}{ll} \therefore & \alpha+\beta=-p, \alpha \beta=q \\ \text { and } & \gamma+\delta=-r, \gamma \delta=s \end{array} $
Now, $(\alpha-\gamma)(\alpha-\delta)(\beta-\gamma)(\beta-\delta)$
$ \begin{aligned} & =\left[\alpha^{2}-(\gamma+\delta) \alpha+\gamma \delta\right]\left[\beta^{2}-(\gamma+\delta) \beta+\gamma \delta\right] \\ & =\left(\alpha^{2}+r \alpha+s\right)\left(\beta^{2}+r \beta+s\right) \\ & =(\alpha \beta)^{2}+r(\alpha+\beta) \alpha \beta+s\left(\alpha^{2}+\beta^{2}\right)+\alpha \beta r^{2}+r s(\alpha+\beta)+s^{2} \\ & =q^{2}-r q p+s\left(p^{2}-2 q\right)+q r^{2}-r s p+s^{2} \\ & =(q-s)^{2}-r q p-r s p+s p^{2}+q r^{2} \end{aligned} $
