The number of real solutions of the equation x(x^2+3|x|+5|x-1|+6|x-2|)=0 is _________.

Question

The number of real solutions of the equation $x({x}^{2}+3|x|+5|x-1|+6|x-2|)=0$ is \_\_\_\_\_\_\_\_\_.

Answer: 1

Accepted Answer

The given equation is $x({x}^{2}+3|x|+5|x-1|+6|x-2|)=0$, which can be solved by analyzing it in parts. It can be broken down into: $x=0$ and ${x}^{2}+3|x|+5|x-1|+6|x-2|=0$. 
For $x=0$, it's clear that it is a solution to the equation since it makes the entire expression equal to zero.
Case (I)

$xCase (II)

$ $$\& 0Case (III)$$

$1Case (IV)

$$$\begin{gathered} x>2 \ {x}^{2}+3x+5(x-1)+6(x-2)=0 \ {x}^{2}+14x-19=0 \end{gathered}$$$

All roots less than 2

$\Rightarrow$ no solution

Here $

Solutions · 1