The sum of the real roots of the equation $x^4-2 x^3+x-380=0$ is
The sum of the real roots of the equation $x^4-2 x^3+x-380=0$ is
$-1$
$0$
$1$
$2$
Solution
Considering the question to be, to find the roots of the equation $x^4-2 x^3+x-380=0$
Now, by trial error method we find that $x=5$ is a solution to the equation
$\Rightarrow \quad 5^4-2 \times 5^3+5-380=0$
$\begin{aligned} & =625-250+5-380=0 \\ & =380-380=0\end{aligned}$
Thus, $x=5$ is a solution.
We also find that $x=-4$ is a solution
$\begin{aligned} & (-4)^4-2 \times(-4)^3-4-380=0 \\ & =256+128-4-380 \\ & =380-380=0\end{aligned}$
The real roots are 5 and -4 . Sum of real roots are $5+(-4)=1$