• # question_answer If the equation ${{x}^{4}}+a{{x}^{3}}-13{{x}^{2}}+bx-4=0$ has one repeated root and one more root being $2+\sqrt{5},$ then A) Repeated root is 2 B) $a=10,\text{ }b=-\,20$ C) $a=0,\text{ }b=-\,20$ D) none of these

 Let $f(x)={{x}^{4}}+a{{x}^{3}}-13{{x}^{2}}+bx-4=0$ If $2+\sqrt{5}$ is one root then other has to be $2-\sqrt{5}$ Let $\pm$ be repeated root, then Product of the roots is $(2+\sqrt{5})(2-\sqrt{5}){{\alpha }^{2}}=-4$$\Rightarrow$${{\alpha }^{2}}=4\operatorname{or}\,\alpha =\pm 2$                                    ...(i) But α can take only one value, so now consider $\sum{\alpha \beta =-13}$
 $\therefore -1+\left( 4-2\sqrt{5} \right)\alpha +\left( 4+2\sqrt{5} \right)\alpha +{{\alpha }^{2}}=-13$ ${{\alpha }^{2}}+8\alpha +12=0$ $\left( \alpha +6 \right)\left( \alpha +2 \right)=0$ $\alpha =-2\,\,or\,\,-6$                                       ...(ii) From result (i) and (ii), $\alpha =-2$ Sum of the roots is, $2+\sqrt{5}+2-\sqrt{5}+\left( 2 \right)+\left( -2 \right)=\operatorname{a}=0$ Similarly $b=-20$