Question

The number of distinct real roots of $x^4-4x^3+12x^2+x-1=0$ is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (A)

$2$

$f\left(x\right)=x^4-4x^3+12x^2+x-1$
Let $f\left(x\right)$ has four distinct real roots
$\Rightarrow f^{{}^{\prime }}\left(x\right)={4x}^3-12x^2+24x+1$
$f^{{}^{\prime }}\left(x\right)$ has three distinct real roots
$f^{{}^{\prime \prime }}\left(x\right)={12x}^2-24x+24=12(x^2-2x+2)$
$D=-4<0$
$f^{{}^{\prime \prime }}\left(x\right)$ cannot have 2 real solutions.
So, f(x) cannot have four real distinct roots
It can have 2 or 0 real roots.
$f\left(0\right)=-1,f\left(1\right)=9$
$\Rightarrow$ At least one real solution
So, 2 real distinct solutions.