Question

If the roots of ax^2 + bx +c = 0 are in the ratio 3:4, prove that 12b^2 = 49ac?

Answer 1

`12b^2 = 49ac`

Explanation:

`y = ax^2 + bx + c = 0`
Reminder of the improved quadratic formula (Socratic Search)
Determinant --> `D = d^2 = b^2 - 4ac`, with `d = +- sqrtD`
The 2 real roots are:
`x = -b/(2a) +- d/(2a)`
`x1 = (-b + d)/(2a)`
`x2 = (-b - d)/(2a) = (-(b + d))/(2a)`
The ration `(x1)/(x2) = 3/4 = (d - b)/-(b + d)`
Cross multiply -->
- 3b - 3d = 4d - 4b
4b - 3b = 4d + 3d
b = 7d.
Square both sides -->
`b^2 = 49d^2 = 49(b^2 - 4ac) = 49b^2 - 196 ac`
`48b^2 = 196ac`. Simplify by 4.
`12b^2 = 49ac`

Answer 2

Let `alpha and beta` are two roots of the given quadratic equation `ax^2+bx+c=0`

Hence `alpha+beta=-b/a and alphabeta=c/a`

Again it is also given that `alpha/beta=3/4`
Let `alpha=3k and beta=4k`

So `7k=-b/a and 12k^2=c/a`

Hence `(49k^2)/(12k^2)=(-b/a)^2/(c/a)^2=b^2/(ac)`

`=>12b^2=49ac`

Proved