Question

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

Answer 1

Let,

$f\left(x\right)=x^4-4x^3+12x^2+x-1$

Consider $f\left(x\right)$ has four distinct roots

$\Rightarrow f^{\prime }\left(x\right)=4x^3-12x^2+24x+1$

$f^{\prime }\left(x\right)$ has three distinct roots

$\begin{array}{c}{f}^{\prime \prime }\left(x\right)=12x^2-24x+24\\ =12\left(x^2-2x+2\right)\end{array}$

$D=-4<0$

Therefore, $f^{\prime \prime }\left(x\right)$ can not have $2$ real solutions.

So, $f\left(x\right)$ can not 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.