Absolute value problem 3 by Dhaval Furia

Let $f(x) = |x|(6x^2+1)-5x^2$ . We note that $f(x)$ is even. Therefore we only need to consider $f(x) = 0$ for $x \ge 0$ . Then we have $f(x) = x(6x^2+1)-5x^2$ and:

$\begin{aligned} x(6x^2 - 5x + 1) & = 0 \\ x(3x-1)(2x-1) & = 0 \\ \implies x & = 0, \frac 13, \frac 12 & \small \blue{\text{for }x \ge 0} \end{aligned}$

Therefore, there are $\boxed 5$ solutions, $x = 0, \pm \dfrac 13, \pm \dfrac 12$ , to the equation.

For $x\geq 0$ , there are three solutions : $0,\dfrac {1}{2},\dfrac {1}{3}$ . For $x<0$ , there are two solutions : $-\dfrac 12,-\dfrac 13$ .

So, in all, there are five solutions .