A) \[f(x)\text{divides}\,\text{g}\,(x)\]
B) \[g(x)=xf(x)\]
C) \[g\,(x)=(x-b-q)f(x)\]
D) None of these
Correct Answer: A
Solution :
[a]Let \[\alpha \in R-Q\]be a common root of \[f(x)=0\]and \[g(x)=0,\]then, \[{{\alpha }^{2}}+a\alpha +b=0\]\[\Rightarrow \]\[{{\alpha }^{2}}=-a\alpha -b\] |
On putting this in \[{{\alpha }^{3}}+p\alpha +q=0,\] we get \[({{a}^{2}}-b+p)\alpha +ab+q=0\] |
As \[\alpha \] is irrational and a, b, p, \[q\in Q.\] |
\[p=b-{{a}^{2}},q=-ab\] |
This give, \[g(x)=(x-a)f(x)\] |
\[\therefore \] \[f(x)\]divides \[g(x).\] |
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