Question

The number of distinct real root(s) of $x^4-4x^3+12x^2+x-1=0$ is

Answer 1

For $f\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+e$
$\underset{x\to \infty }{lim}f\left(x\right)=\underset{x\to -\infty }{lim}f\left(x\right)=\infty$
$f\left(x\right)$ will have at maximum $4$ real roots.
Possibilities of $f\left(x\right)$ are


or


or


or



$f\left(x\right)=x^4-4x^3+12x^2+x-1$
$\Rightarrow f^{\prime }\left(x\right)=4x^3-12x^2+24x+1$
$\Rightarrow f^{\prime \prime }\left(x\right)=12x^2-24x+24=12(x^2-2x+2)$
$\Rightarrow f^{\prime \prime }\left(x\right)=12\left(\right(x-1{)}^2+1)>0$
$\Rightarrow f^{\prime }\left(x\right)$ is increasing.
$f^{\prime }\left(x\right)=0$ will have only one real root.
$f^{\prime }\left(x\right)=4x^3-12x^2+24x+1$
$f^{\prime }\left(0\right)=1$


$f^{\prime }\left(x\right)$ will have only one root that too negative.
Also, $f\left(0\right)=-1$


So, $f\left(x\right)$ will have $2$ distinct real roots.


Alternate Solution:
$f\left(x\right)=x^4-4x^3+12x^2+x-1$
$\Rightarrow f^{\prime }\left(x\right)=4x^3-12x^2+24x+1$
$\Rightarrow f^{\prime \prime }\left(x\right)=12x^2-24x+24=12(x^2-2x+2)$
$\Rightarrow f^{\prime \prime }\left(x\right)=12\left(\right(x-1{)}^2+1)>0$
$\Rightarrow f^{\prime }\left(x\right)$ is increasing.
$\Rightarrow f^{\prime }\left(x\right)=0$ will have only one real root.
Hence, $f\left(x\right)=0$ has maximum $2$ distinct real roots.

We know that, complex roots occur in conjugate pair when coefficients are real and $f\left(0\right)=-1$.
So, $f\left(x\right)$ will have $2$ distinct real roots.