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: A
Solution :
Given that, \[{{z}^{2}}+(p+iq)z+r+is=0\] ... (i) Let \[Z=\alpha \] (where, \[\alpha \] is real) be a root of Eq. (i), then \[{{\alpha }^{2}}+(p+iq)\alpha +r+is=0\] \[\Rightarrow \] \[{{\alpha }^{2}}+p\alpha +r+i(q\alpha +s)=0\] On equating real and imaginary parts, we get \[{{\alpha }^{2}}+p\alpha +r=0\] ... (ii) and \[q\alpha +s=0\Rightarrow \alpha =\frac{-s}{q}\] On putting the value of \[\alpha \] in Eq. (ii), we get \[{{\left( \frac{-s}{q} \right)}^{2}}+\left( \frac{-s}{q} \right)p+r=0\] \[\Rightarrow \] \[{{s}^{2}}-pqs+{{q}^{2}}r=0\] \[\Rightarrow \] \[pqs={{s}^{2}}+{{q}^{2}}r\]
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