Question

The value of a for which the equations ${{x}^{3}}+ax+1=0$ and ${{x}^{4}}+a{{x}^{2}}+1=0$ have a common root is
A) a=2
B) a=-2
C) a=0
D) none of these

Answer 1

Hint: In this question, we are given the two equations therefore we should try to construct a new equation which would be satisfied only if both the equations are satisfied. Solving that equation, we should try to calculate the solutions of that new equation and from that we can obtain the value of a.

Complete step-by-step answer:
The equations given in the question are
${{x}^{3}}+ax+1=0.............(1.1)$
and
${{x}^{4}}+a{{x}^{2}}+1=0.................(1.2)$
Now, if we multiply equation (1.1) with x, we get
${{x}^{4}}+a{{x}^{2}}+x=0.............(1.3)$
Now, if $\alpha $ is a common root of (1.1) and (1.2), it must also satisfy equation (1.3), therefore, we should have
${{\alpha }^{4}}+a{{\alpha }^{2}}+1=0...............(1.4)$ and
\[{{\alpha }^{4}}+a{{\alpha }^{2}}+\alpha =0.............(1.5)\]
Therefore, subtracting equation (1.4) from equation (1.5), we obtain
\[\begin{align}
  & {{\alpha }^{4}}+a{{\alpha }^{2}}+\alpha -\left( {{\alpha }^{4}}+a{{\alpha }^{2}}+1 \right)=0 \\
 & \Rightarrow \alpha -1=0 \\
 & \Rightarrow \alpha =1 \\
\end{align}\]
Thus, 1 should be a common root to both the equations, thus it should also satisfy equation (1.1), using x=1 in equation (1.1), we obtain
$\begin{align}
  & {{1}^{3}}+(a)1+1=0 \\
 & \Rightarrow 2+a=0 \\
 & \Rightarrow a=-2.............(1.6) \\
\end{align}$
Thus, from equation (1.6), we obtain the answer to be a=-2 which matches option (B) in the question, hence option (B) is the correct answer to this question.

Note: To obtain the value of a, we could also have used the fact that 1 should be a root of equation (1.2) also, therefore we would have obtained ${{1}^{4}}+a{{1}^{2}}+1=0\Rightarrow a=-2$ which is the same answer as obtained in the solution above.