Question

lf $x^3+ax+1=0$ and $x^4+a{x}^2+1=0$ have a common root then the exhaustive set of values of a is

• A. $(-\infty ,-2)$
• B. $[-2,\infty )$
• C. $\{-2\}$
• D. $[-2,2]$

Answer 1

Answer: (C)

$\{-2\}$

Given $x^3+ax+1=0,x^4+a{x}^2+1=0$

Let $\alpha$ be the common root.

$\Rightarrow {\alpha }^3+a\alpha +1=0\Rightarrow {\alpha }^3+a\alpha =-1$

${\alpha }^4+a{\alpha }^2+1=0$

$\alpha ({\alpha }^3+a\alpha )+1=0$

$\alpha (-1)+1=0$

$\alpha =1$

$(1{)}^3+a(1)+1=0$

$a=-2$

values of $a=-2$ alone