A) \[{{x}^{2}}+abx+{{b}^{3}}=0\]
B) \[{{x}^{2}}-abx+{{b}^{3}}=0\]
C) \[b{{x}^{2}}+x+a=0\]
D) \[{{x}^{2}}+ax+ab=0\]
Correct Answer: A
Solution :
\[f(x)={{x}^{2}}+ax+b=0\]Þ \[p+q=-a\]and\[pq=b\] Now required equation whose roots are \[{{p}^{2}}q\]and \[p{{q}^{2}}\] Therefore sum of roots \[={{p}^{2}}q+p{{q}^{2}}=pq(p+q)=-ab\] and product of roots =\[p{{q}^{2}}.q{{p}^{2}}={{(pq)}^{3}}={{b}^{3}}\] Thus equation is\[{{x}^{2}}+abx+{{b}^{3}}=0\].You need to login to perform this action.
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