Question

The number of values of $a$ for which equation $x^3+ax+1=0$ and $x^4+a{x}^2+1=0$ have a common root, is:

• A. $0$
• B. $1$
• C. $2$
• D. $\infty$

Answer 1

The correct option is B $1$

$x^3+ax+1=0$ and $x^4+a{x}^2+1=0$
Subtracting (i) from (ii), we get
$x^4-x^3+a(x^2-x)=0$
$x^3(x-1)+ax(x-1)=0$
$(x^3+ax)(x-1)=0$
$x(x^2+a)(x-1)=0$
$x=1$ or $x=0$ are common roots and $x^2+a=0$ is not possible for real values of '$a$'.
Substituting $x=1$ in the equation (1), we get
$1+a+1=0$
$x=-2$
Substituting in equation (2), we get
$x^4-2x^2+1=0$
$(x^2-1{)}^2=0$
$x=\pm 1$
Hence, common root is $1$ and there can be only one value of '$a$',
ie, $a=-2$.