A) \[\frac{a}{p}\]
B) \[1\]
C) \[{{\left( \frac{a}{p} \right)}^{2}}\]
D) \[0\]
E) \[{{\left( \frac{p}{a} \right)}^{2}}\]
Correct Answer: C
Solution :
Since,\[\alpha ,\beta \]are the roots of the equation \[ax+bx+c=0\] Then, \[\alpha =\frac{-b}{2a}+\frac{\sqrt{{{b}^{2}}-4ac}}{2a},\] \[\beta =\frac{-b}{2a}-\frac{\sqrt{{{b}^{2}}-4ac}}{2a},\] ??(i) And\[\alpha +k,\beta +k\]are the roots of the equation \[p{{x}^{2}}+qx+r=0\] then, \[\alpha +k=-\frac{q}{2p}+\frac{\sqrt{{{q}^{2}}-4pr}}{2p}\] and \[\beta +k=-\frac{q}{2p}-\frac{\sqrt{{{q}^{2}}-4pr}}{2p}\] \[\Rightarrow \]\[k=-\frac{q}{2p}+\frac{\sqrt{{{q}^{2}}-4pr}}{2p}+\frac{b}{2a}-\frac{\sqrt{{{b}^{2}}-4ac}}{2a}\] [ from (i)] and \[k=-\frac{q}{2p}-\frac{\sqrt{{{q}^{2}}-4pr}}{2p}+\frac{b}{2a}-\frac{\sqrt{{{b}^{2}}-4ac}}{2a}\] [from (i)] \[\Rightarrow \] \[\frac{\sqrt{{{q}^{2}}-4pr}}{2p}-\frac{\sqrt{{{b}^{2}}-4ac}}{2a}\] \[=\frac{-\sqrt{{{q}^{2}}-4pr}}{2p}+\frac{\sqrt{{{b}^{2}}-4ac}}{2a}\] \[\Rightarrow \] \[\frac{\sqrt{{{q}^{2}}-4pr}}{p}=\frac{\sqrt{{{b}^{2}}-4ac}}{a}\] \[\Rightarrow \] \[\frac{{{q}^{2}}-4pr}{{{p}^{2}}}=\frac{{{b}^{2}}-4ac}{{{a}^{2}}}\] \[\therefore \] \[\frac{{{b}^{2}}-4ac}{{{q}^{2}}-4pr}={{\left( \frac{a}{p} \right)}^{2}}\]
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