A) \[{{p}^{3}}-(3p-1)q+{{q}^{2}}=0\]
B) \[{{p}^{3}}-q(3p+1)+{{q}^{2}}=0\]
C) \[{{p}^{3}}+q(3p-1)+{{q}^{2}}=0\]
D) \[{{p}^{3}}+q(3p+1)+{{q}^{2}}=0\]
Correct Answer: A
Solution :
The equation is \[{{x}^{2}}+px+q=0\] Let \[\alpha \] be one of the root, then as per problem, second root is \[{{\alpha }^{2}}\]. From the principle of quadratic equation. \[{{\alpha }^{2}}+\alpha =-p\] ....(1) and \[{{\alpha }^{3}}=q\] ...(2) From eq (1)+ eq(2): \[{{\alpha }^{3}}+{{\alpha }^{2}}+\alpha =q-p\] \[\Rightarrow \,\,\alpha ({{\alpha }^{2}}+\alpha +1)=q-p\] \[\Rightarrow \,\,\alpha (-p+1)=q-p\][since \[{{\alpha }^{2}}+\alpha =-p\] from \[e{{q}^{n}}\](1)] \[\Rightarrow \,\,\alpha =\frac{q-p}{1-p}=\frac{p-q}{p-1}\] Putting this value of a in equation (1) \[{{\left( \frac{p-q}{p-1} \right)}^{2}}+\left( \frac{p-q}{p-1} \right)=-p\] \[\Rightarrow \,\,\frac{{{p}^{2}}-2pq+{{q}^{2}}}{{{(p-1)}^{2}}}+\frac{p-q}{(p-1)}=-p\] \[\Rightarrow \,\,\frac{{{p}^{2}}-2pq+{{q}^{2}}+(p-1)(p-q)}{{{(p-1)}^{2}}}=-p\] \[\Rightarrow \,{{p}^{2}}-2pq+{{q}^{2}}+{{p}^{2}}-pq-p+q=-p({{p}^{2}}-2p+1)\]\[\Rightarrow 2{{p}^{2}}-3pq+{{q}^{2}}-p+q=-{{p}^{3}}+2{{p}^{2}}-p.\] \[\Rightarrow {{p}^{3}}-3pq+q+{{q}^{2}}=0\] \[\Rightarrow {{p}^{3}}-q(3p-1)+{{q}^{2}}=0\]
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