A) \[{{r}^{2}}b=q{{c}^{2}}\]
B) \[r{{b}^{2}}=c{{q}^{2}}\]
C) \[{{r}^{2}}c=q{{b}^{2}}\]
D) \[r{{c}^{2}}=b{{q}^{2}}\]
E) \[{{r}^{2}}q={{c}^{2}}b\]
Correct Answer: B
Solution :
Let\[\alpha ,\beta \]be the roots of\[{{x}^{2}}+bx+c=0\]and \[\alpha ,\beta \]are 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\[\frac{\alpha }{\beta }=\frac{\alpha }{\beta }\] \[\Rightarrow \] \[\frac{\alpha +\beta }{\alpha -\beta }=\frac{\alpha +\beta }{\alpha -\beta }\] \[\Rightarrow \] \[\frac{{{(\alpha +\beta )}^{2}}}{{{(\alpha -\beta )}^{2}}}=\frac{{{(\alpha +\beta )}^{2}}}{{{(\alpha -\beta )}^{2}}}\] \[\Rightarrow \] \[\frac{{{b}^{2}}}{{{b}^{2}}-4c}=\frac{{{q}^{2}}}{{{q}^{2}}-4r}\] \[\Rightarrow \] \[{{b}^{2}}r={{q}^{2}}c\]
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