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

2 Answers
Jan 10, 2017

Answer:

#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#

Jan 10, 2017

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