Question

The quadratic eqn x²-6x+a=0 and x²-cx+6=0 have one root in common. The other roots of the first and second eqn are integers in the ratio 4:3. Then the common roots is-

Answer 1

let the roots of the first be s and 4r
let the roots of the second be s and 3r

1st: x^2 - 6x + a = 0
sum of roots = 6
product of roots = a
s+4r = 6 and 4rs = a

2nd: x^2 - cx + 6 = 0
sum of roots = c
product of roots = 6
s+3r =c and 3rs = 6 ---> rs = 2

so in 4rs = a , we have a = 8

so now we have the first equation as
x^2 - 6x + 8 = 0
(x-2)(x-4) = 0
x = 2 or x = 4

so for the product of the roots of the second to be 6
they must be 2 and 3, with 2 being the common one

check:
1st: x^2- 6x + 8 = 0 --->roots are 2,4
2nd: x^2 - 5x + 6 = 0 --->roots are 2,3

2 is common, the others are 4 and 3 which is in ratio of 4:3