A) 4
B) 1
C) 0
D) -4
Correct Answer: D
Solution :
Let the roots are \[\alpha ,\beta \] of \[{{x}^{2}}-bx+c=0\]and \[{\alpha }',{\beta }'\]be roots of \[{{x}^{2}}-cx+b=0\] Now \[\alpha -\beta =\sqrt{{{(\alpha +\beta )}^{2}}-4\alpha \beta }=\sqrt{{{b}^{2}}-4c}\] .....(i) and \[\alpha '-\beta '=\sqrt{{{(\alpha '+\beta ')}^{2}}-4\alpha '\beta '}=\sqrt{{{c}^{2}}-4b}\] .....(ii) But \[\alpha -\beta =\alpha '-\beta '\] Þ \[\sqrt{{{b}^{2}}-4c}=\sqrt{{{c}^{2}}-4b}\,\,\Rightarrow \,\,{{b}^{2}}-4c={{c}^{2}}-4b\] Þ \[{{b}^{2}}-{{c}^{2}}=4c-4b\] Þ \[(b+c)(b-c)=4(c-b)\]Þ\[b+c=-4\]
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