The distance between the two points (x_1,y_1) and (x_2,y_2) is given as d=sqrt((x_2-x_1)^2+(y_2-y_1)^2).
For this problem, it is given that (x_1,y_1)=(5,8), (x_2,y_2)=(a,2), and d=3sqrt5. We need to solve for a by plugging these values into the distance formula.
3sqrt5=sqrt((a-5)^2+(2-8)^2).
To solve for a, begin by squaring both sides of the equation and simplify a bit.
(3sqrt5)^color(green)2=sqrt((a-5)^2+(2-8)^2)^color(green)2
3^2sqrt5^2=sqrt((a-5)^2+(2-8)^2)^2
9*5=(a-5)^2+(2-8)^2
45=(a-5)^2+("-"6)^2
45=a^2-10a+25+36
45=a^2-10a+61
Now subtract 45 from both sides to get an equation in quadratic form.
45color(green)(-45)=a^2-10a+61color(green)(-45)
0=a^2-10a+16
Finally, solve for a using either the quadratic formula or the factoring method. I'll be factoring.
0=a^2-2a-8a+16
0=a(a-2)-8(a-2)
0=(a-8)(a-2)
a=8or2
Therefore the possible values for a are defined by the set {a:a=2,8}
