Question

Number of non-real roots of the equation $x^4-4x-1=0$ is

• A. 2

Answer 1

The correct option is A 2
Let $f\left(x\right)=x^4-4x-1=0$
The sign changes of the coefficients are $+,-,-$
according to Descartes' rule,
$1$ change of sign $\Rightarrow$ maximum 1 positive roots.

Now $f(-x)=x^4+4x-1=0$
The sign changes of the coefficients are $+,+,-$
according to Descartes' rule,
$1$ change of sign $\Rightarrow$ maximum 1 negative root.

Hence the total maximum number of real roots $=1+1=2$
Since total number of roots of an equation of order $4$ is $4$, hence the number of non-real roots of the equation $4-2=2$