• # question_answer If the quadratic equations ${{x}^{2}}+ax+b=0$and ${{\mathbf{x}}^{2}}+\mathbf{bx}+\mathbf{a}=\mathbf{0}$have a common root then the numerical value of $\left( \mathbf{a}+\mathbf{b} \right)$ is: A)  1                    B)  $-1$              C)  0                                 D)  2

[b] $\because \alpha$be the comment roots of the given quadratic equation we have  $\therefore {{\alpha }^{2}}+a.\alpha +b=0$                 ??..(1) $\And {{\alpha }^{2}}+b.\alpha +a=0$                       ??  (2) (1) ?(2), we have $\alpha (a-b)+b-a=0$ $\alpha (a-b)=\left( a-b \right)$ $\therefore \alpha =1$ Putting these value in the given quadratic equation we have. ${{1}^{2}}+\alpha .1+b=0$ $\Rightarrow a+b=-1$ Hence option [b] is correct.