if one of the root of the equation ax^2+bx+c=0 is four times the other show that 4b^2-25ac=0...

Question

help please show working plz

Answer 1

(x-q)(x-4q) = a x^2 + b x + c

x^2 - 5qx + 4 q^2 =a x^2 + bx + c
if a = 1
b = -5 q
c = 4 q^2

then
4 b^2 = 100 q^2
and
25 a c = 100 q^2
and
100 q^2 -100 q^2 = 0 all right