use the distance formula
a point on the x axis is where y=0
disatnce formula
distance between (x1,y1) and (x2,y2) is
[tex]D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
we have the point (4,3) and (x,0)
and D=6
[tex]6=\sqrt{(x-4)^2+(0-3)^2}[/tex]
[tex]6=\sqrt{x^2-8x+16+(-3)^2}[/tex]
[tex]6=\sqrt{x^2-8x+16+9}[/tex]
[tex]6=\sqrt{x^2-8x+25}[/tex]
square both sides
[tex]36=x^2-8x+25[/tex]
minus 36 both sides
0=x²-8x-11
factor
we can't
use quadratic formula
for
0=ax²+bx+c
[tex]x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}[/tex]
so
0=1x²-8x-11
a=1
b=-8
c=-11
[tex]x=\frac{-(-8)+/-\sqrt{(-8)^2-4(1)(-11)}}{2(1)}[/tex]
[tex]x=\frac{8+/-\sqrt{64+44}}{2}[/tex]
[tex]x=\frac{8+/-\sqrt{108}}{2}[/tex]
[tex]x=\frac{8+/-6\sqrt{3}}{2}[/tex]
[tex]x=4+/-3\sqrt{3}[/tex]
the 2 points are
[tex](4+3\sqrt{3},0) \space\ and \space\ (4-3\sqrt{3},0)[/tex]
