Answer
$\left( 4+3\sqrt{3},\,0 \right)$ and $\left( 4-3\sqrt{3},\,0 \right)$
Work Step by Step
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 $\left( 4-3\sqrt{3},\,0 \right)$.
