Theory of Equations 1 Question 45
46. If $\alpha, \beta$ are the roots of $a x^{2}+b x+c=0,(a \neq 0)$ and $\alpha+\delta, \beta+\delta$ are the roots of $A x^{2}+B x+C=0,(A \neq 0)$ for some constant $\delta$, then prove that
$ \frac{b^{2}-4 a c}{a^{2}}=\frac{B^{2}-4 A C}{A^{2}} $
$(2000,4 M)$
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Solution:
- Since, $\alpha+\beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}$
$ \text { and } \quad \alpha+\delta+\beta+\delta=-\frac{B}{A},(\alpha+\delta)(\beta+\delta)=\frac{C}{A} $
$ \text { Now, } \quad \alpha-\beta=(\alpha+\delta)-(\beta+\delta) $
$ \begin{array}{ll} \Rightarrow & (\alpha-\beta)^{2}=[(\alpha+\delta)-(\beta+\delta)]^{2} \\ \Rightarrow & (\alpha+\beta)^{2}-4 \alpha \beta=\left(\overline{\alpha+\delta}+\overline{\beta+\delta)^{2}-4(\alpha+\delta)(\beta+\delta)}\right. \\ \Rightarrow & -\frac{b}{a}^{2}-\frac{4 c}{a}=-\frac{B^{2}}{A}-\frac{4 C}{A} \\ \Rightarrow & \frac{b^{2}}{a^{2}}-\frac{4 c}{a}=\frac{B^{2}}{A^{2}}-\frac{4 C}{A} \\ \Rightarrow & \frac{b^{2}-4 a c}{a^{2}}=\frac{B^{2}-4 A C}{A^{2}} \end{array} $
