## Precalculus (10th Edition)

$\left( 4+3\sqrt{3},\,0 \right)$ and $\left( 4-3\sqrt{3},\,0 \right)$
The points on the $x-$axis are represented by the point $\left( x,0 \right)$. By using the distance formula with ${{P}_{1}}=\left( {{x}_{1}},{{y}_{1}} \right)=\left( 4,-3 \right)$ , ${{P}_{2}}=\left( {{x}_{2}},{{y}_{2}} \right)=\left( x,0 \right)$ and $d\left( {{P}_{1}},\,{{P}_{2}} \right)=6$, $6=\sqrt{{{\left( x-4 \right)}^{2}}+{{\left( 0-\left( -3 \right) \right)}^{2}}}$ By simplifying, $6=\sqrt{{{\left( x-4 \right)}^{2}}+{{\left( 3 \right)}^{2}}}$ Solving for $x$, $6=\sqrt{{{\left( x-4 \right)}^{2}}+9}$ Squaring on both sides, $36={{\left( x-4 \right)}^{2}}+9$ $36={{\left( x-4 \right)}^{2}}+9$ Subtract $9$on both sides, $36-9={{\left( x-4 \right)}^{2}}+9-9$ $27={{\left( x-4 \right)}^{2}}$ $\Rightarrow {{\left( x-4 \right)}^{2}}=27$ By square root method, $\Rightarrow x-4=\pm \sqrt{27}$ $\Rightarrow x-4=\pm \,3\sqrt{3}$ Add $4$ on both sides, $\Rightarrow x=4\pm \,3\sqrt{3}$ $\Rightarrow x=4+3\sqrt{3}$ and $x=4-3\sqrt{3}$ Therefore, the points are $\left( 4+3\sqrt{3},\,0 \right)$ and $\left( 4-3\sqrt{3},\,0 \right)$.