Question

The number of positive real roots of $x^4-4x-1=0$ is

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

Answer: (C)

$1$

The given function $x^4-4x-1=0$ has only one change in sign.

From descartes rules of sign, there is a sign change for $x^4$ to $x$ from positive to negative. Therefore, the given equation has at most one positive real root.

Now, the function $f(-x)=x^4+4x-1$ also has only one change in sign.

From descartes rules of sign, there is a sign change for $4x$ to $1$ from positive to negative. Therefore, the given equation has at most one negative real root.

Also, we have

$f(-\infty )=+ve,f\left(\infty \right)=+ve,f\left(0\right)=-1$

Therefore, the equation $x^4-4x-1=0$ has one negative real root and one positive real root.

Hence, the number of positive real roots of $x^4-4x-1=0$ is $1$.