The correct option is C 2
Consider the first equation.
Let the roots be α,β.
Then the equation becomes
x2−(α+β)x+αβ=0.
Comparing with the given equation
x2−6x+a=0 we get
αβ=a and α+β=6 ...(i)
Similarly let the roots of the second equation be α,δ.
Hence
x2−(α+δ)+α.δ=0
Comparing with the given equation x2−cx+6=0
Hence
α+δ=c and α.δ=6 ...(ii)
Now from i and ii we know that
α.β=a and α.δ=6
α.βα.δ=a6
βδ=a6.
Now it is given that the other two roots of the equations (roots excluding the common root) are in the ratio of 4:3.
Hence
βδ=43
Or
a6=43
a=8
Hence the first equation becomes
x2−6x+8=0
(x−4)(x−2)=0
x=2,4.
Now out of 2 and 4, any one of the following can be the root. The product of the roots of the second equation is 6. If 4 is taken as the common root then the other root becomes a fraction. However it is clearly stated that the roots are integral. Hence the possible common root is 2.
Thus α=2.
And
α.δ=6
Hence
δ=3.
Therefore the equation becomes
x2−(3+2)x+6=0
x2−5x+6=0.
Hence
a=8,c=5 and common root is 2.