• question_answer Let $a,b,p,q\in Q$ and suppose that $f(x)={{x}^{2}}+ax+b=0$ and $g(x)={{x}^{3}}+px+q=0$ have a common irrational roots. Then, 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

[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 $g(x).$