Question

Find all points on the x-axis that are a distance 5 from P(-8,3)

Answer 1

On the x axis means that the y value is 0 or the point is (x,0)

distance formula

D=$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
we know that distance is 5
the first point is (x1,y1)
the (x2,y2) is (x,0)

so
D=5
x1=-8
y1=3
x2=x
y2=0

5=$\sqrt{(-8-x)^2+(3-0)^2}$
5=$\sqrt{(-8-x)^2+(3)^2}$
5=$\sqrt{64+16x+x^2+9}$
5=$\sqrt{x^2+16x+73}$
square both sides
25=${x^2+16x+73}$
minus 25 both sides
0=x^2+16x+48
factor
0=(x+4)(x+12)
set each to zero

x+4=0
x=-4

x+12=0
x=-12

the points are

(-4,0) and (-12,0)