A) \[{{r}^{2}}c={{b}^{2}}q\]
B) \[{{r}^{2}}b={{c}^{2}}q\]
C) \[r{{b}^{2}}=c{{q}^{2}}\]
D) \[r{{c}^{2}}=b{{q}^{2}}\]
Correct Answer: C
Solution :
Let \[\alpha ,\beta \] be the roots of\[{{x}^{2}}+bx+c=0\]and \[\alpha ',\beta '\] be the roots of \[{{x}^{2}}+qx+r=0\]. Then \[\alpha +\beta =-b,\alpha \beta =c,\alpha '+\beta '=-q,\alpha '\beta '=r\] It is given that \[\frac{\alpha }{\beta }=\frac{\alpha '}{\beta '}\Rightarrow \frac{\alpha +\beta }{\alpha -\beta }=\frac{\alpha '+\beta '}{\alpha '-\beta '}\] Þ \[\frac{{{(\alpha +\beta )}^{2}}}{{{(\alpha -\beta )}^{2}}}=\frac{{{(\alpha '+\beta ')}^{2}}}{{{(\alpha '-\beta ')}^{2}}}\,\,\,\Rightarrow \frac{{{b}^{2}}}{{{b}^{2}}-4c}=\frac{{{q}^{2}}}{{{q}^{2}}-4r}\] Þ \[{{b}^{2}}r={{q}^{2}}c\]
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