## Fundamentals of Physics Extended (10th Edition)

$x_{1}=2-\sqrt {13}\approx -1,606m$
Let's assume that our starting point is $x=0$ so the work done by conservative force at any given $x$ will be $W=\int ^{x}_{0}F\left( x\right) dx=\int ^{x}_{0}\left( 6x-12\right) =3x^{2}-12x(1)$ So the change in potential energy for conservative force will be: $\Delta U=-W\left( 2\right)$ So from (1) and (2) we get $\Delta U=-W=12x-3x^{2}(3)$ And also $\Delta U=U\left( x\right) -U_{0}\left( 4\right)$ From (3) and (4) we get: $U\left( x\right) =U_{0}+12x-3x^{2}(5)$ $12x-3x^{2}=3\left( 4x-x^{2}\right) =-3\left( x^{2}-4x\right) =-3\left( x^{2}-4x+4\right) +12=12-3\left( x-2\right) ^{2}\left( 6\right)$ From (5) and (6) we get $U\left( x\right) =U_{0}+12-3\left( x-2\right) ^{2}=27+12-3\left( x-2\right) ^{2}=39-3\left( x-2\right) ^{2}=3\left( 13-\left( x-2\right) ^{2}\right)$ So if $U_{x}=0$ then $U\left( x\right) =3\left( 13-\left( x-2\right) ^{2}\right) =0\Rightarrow \left( 13-\left( x-2\right) ^{2}\right) =0\Rightarrow x=2\pm \sqrt {13}$ question asked negative x so $x_{1}=2-\sqrt {13}\approx -1,606m$