The quadratic equations `x^2""-6x""+""a""=""0""a n d ""x^2""-c x""+""6""=""0` have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is

Options

A. both roots more than `alpha`
B. both roots less than `alpha`
C. one root more than `alpha` and other less than `alpha`
D. Can't say anything

Answer

3

Explanation

Let equation `x^(2) - 6x + a = 0` have roots `alpha and beta`.
Also, let equation `x^(2) - cx + 6 = 0` have roots `alpha and gamma`.
Given `beta/gamma - 4/3 rArr (alphabeta)/(alphagamma) = 4/3 rArr a/6 = 4/3 rArr a = 8`
`therefore` First equation is `x^(2) - 6x + 8 = 0`
`rArr (x - 2)(x - 4) = 0`
`rArr x = 2, 4`
If `beta = 4,` then `gamma = (3beta)/(4) = 3`.
If `beta = 2,` then `gamma = (3beta)/(4) = 3/2`, which is not an integer.
Now, `x^(2) - 3x - 4 = 0`
`rArr (x - 4)(x + 1) = 0`
`rArr x = -1, 4`
Clearly, one root is less then 2 and other is more than 2.