If one of the root of the quadratic equation ax^2 + bx +c =0 is three times the other, show that 3b^2 = 16ac

Answer 1

ax^2 + bx + c = 0
x^2 + (b/a)x + (c/a) = 0
For quadratic equations with this form, the sum & product of roots are:
sum = b/a
product = c/a

Let R = one root
Let 3R = the other root (according to the problem)
Substituting to sum & product equations,
R + 3R = b/a
R(3R) = c/a
from the sum,
4R = b/a
R = b/(4a)
substituting this to the product,
R(3R) = c/a
3R^2 = c/a
3(b/(4a))^2 = c/a
3b^2 / (16a^2) = c/a
3b^2 = (16a^2)c/a
3b^2 = 16ac

Hope this helps~ :)

Answer 2

*sorry I mean sum = -b/a, and thus R = -b/(4a) . But the answer's still the same since it is squared.

Answer 3

how to find the roots of the equation