A) \[{{b}^{2}}-4ac\]
B) \[2{{b}^{2}}-ac\]
C) \[{{b}^{2}}-ac\]
D) None of the above
Correct Answer: A
Solution :
Given, roots of the equation\[a{{x}^{2}}+bx+c=0\]are \[\frac{\alpha }{\alpha -1}\]and \[\frac{\alpha +1}{\alpha },\]which are real. \[\therefore \] \[\frac{\alpha }{\alpha +1}+\frac{\alpha +1}{\alpha }=\frac{-b}{a}\] ...(i) and \[\frac{\alpha }{\alpha -1}.\frac{(\alpha +1)}{\alpha }=\frac{x}{a}\] \[\Rightarrow \] \[\frac{\alpha +1}{\alpha -1}=\frac{c}{a}\] \[\Rightarrow \] \[a\alpha +a=c\alpha -c\] \[\Rightarrow \] \[(a-c)\alpha =-c-a\] \[\Rightarrow \] \[\alpha =-\frac{(c+a)}{(a-c)}\] \[\Rightarrow \] \[\alpha =\frac{c+a}{c-a}\] ?(ii) From Eq. (i), \[\frac{\alpha }{\alpha -1}+\frac{\alpha +1}{\alpha }=\frac{-b}{a}\] \[\Rightarrow \] \[\frac{{{\alpha }^{2}}+{{\alpha }^{2}}-1}{\alpha (\alpha -1)}=\frac{-b}{a}\] \[\Rightarrow \] \[\frac{2{{\alpha }^{2}}-1}{\alpha (\alpha -1)}=\frac{-b}{a}\] On putting the value of a from Eq. (i), we get \[\frac{2\left( \frac{c+a}{c-a} \right)-1}{\left( \frac{c+a}{c-a} \right)\left( \frac{c+a}{c-a}-1 \right)}=\frac{-b}{a}\] \[\Rightarrow \] \[\frac{2{{(c+a)}^{2}}-{{(c-a)}^{2}}}{(c+a)(c+a-c+a)}=\frac{-b}{a}\] \[\Rightarrow \]\[\frac{2{{c}^{2}}+2{{a}^{2}}+4ac-{{c}^{2}}-{{a}^{2}}+2ac}{2ac+2{{a}^{2}}}=\frac{-b}{a}\] \[\Rightarrow \] \[\frac{{{c}^{2}}+{{a}^{2}}+6ac}{(2c+2a)a}=\frac{-b}{a}\] \[\Rightarrow \]\[{{c}^{2}}+{{a}^{2}}+6ac=-2bc-2ab\] \[\Rightarrow \]\[{{a}^{2}}+{{c}^{2}}+2ac+2bc+2ab+{{b}^{2}}={{b}^{2}}-4ac\] \[\Rightarrow \] \[{{(a+b+c)}^{2}}={{b}^{2}}-4ac\]
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