A) \[\frac{2b}{ac}=\frac{{{q}^{2}}}{pr}\]
B) \[\frac{b}{ac}=\frac{q}{pr}\]
C) \[\frac{{{b}^{2}}}{ac}=\frac{{{q}^{2}}}{pr}\]
D) None of these
Correct Answer: C
Solution :
Let\[\alpha ,\beta \]be the roots of the equation \[a{{x}^{2}}+2bx+c=0\]and\[y,\delta \]be the roots of the equation\[p{{x}^{2}}+2qx+r=0\]be Also, given \[\frac{\alpha }{\beta }=\frac{\gamma }{\delta }\Rightarrow \frac{\alpha }{\gamma }=\frac{\beta }{\delta }\] \[\Rightarrow \] \[\frac{\alpha +\beta }{\gamma +\delta }=\sqrt{\frac{\alpha \beta }{\gamma \delta }}\] \[\Rightarrow \] \[\frac{-2b/a}{-2q/p}=\sqrt{\frac{c/a}{r/p}}\] \[\Rightarrow \] \[\frac{{{b}^{2}}}{ac}=\frac{{{q}^{2}}}{pr}\]
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