A) \[pqr={{r}^{2}}+{{p}^{2}}s\]
B) \[prs={{q}^{2}}+{{r}^{2}}p\]
C) \[qrs={{p}^{2}}+{{s}^{2}}q\]
D) \[pqs={{s}^{2}}+{{q}^{2}}r\]
Correct Answer: D
Solution :
Given that \[{{z}^{2}}+(p+iq)z+r+is=0\] ......(i) Let \[z=\alpha \] (where \[\alpha \] is real) be a root of (i), then \[{{\alpha }^{2}}+(p+iq)\alpha +r+\]is =0 or \[{{\alpha }^{2}}+p\alpha +r+i(q\alpha +s)\]=0 Equating real and imaginary parts, we have \[{{\alpha }^{2}}+p\alpha +r=0\] and \[q\alpha +s=0\] Eliminating \[\alpha ,\]we get \[{{\left( \frac{-s}{q} \right)}^{2}}+p\left( \frac{-s}{q} \right)+r=0\] or \[{{s}^{2}}-pqs+{{q}^{2}}r=0\] or \[pqs={{s}^{2}}+{{q}^{2}}r\]
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