Question

Find all points on the x axis that are 6 units from the point 4 3

Answer 1

Use the distance formula

a point on the x axis is where y=0

disatnce formula
distance between (x1,y1) and (x2,y2) is
$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

we have the point (4,3) and (x,0)
and D=6

$6=\sqrt{(x-4)^2+(0-3)^2}$
$6=\sqrt{x^2-8x+16+(-3)^2}$
$6=\sqrt{x^2-8x+16+9}$
$6=\sqrt{x^2-8x+25}$
square both sides
$36=x^2-8x+25$
minus 36 both sides
0=x²-8x-11
factor
we can't
use quadratic formula
for
0=ax²+bx+c
$x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}$

so
0=1x²-8x-11
a=1
b=-8
c=-11

$x=\frac{-(-8)+/-\sqrt{(-8)^2-4(1)(-11)}}{2(1)}$
$x=\frac{8+/-\sqrt{64+44}}{2}$
$x=\frac{8+/-\sqrt{108}}{2}$
$x=\frac{8+/-6\sqrt{3}}{2}$
$x=4+/-3\sqrt{3}$

the 2 points are
$(4+3\sqrt{3},0) \space\ and \space\ (4-3\sqrt{3},0)$