Question

The quadratic equations $x^2-6x+a=0$ and $x^2-cx+6=0$ have one root in common. The other roots of the first and second equations are integers in the ratio $4:3$ respectively. Then, the common root is

• A. $3$
• B. $2$
• C. $1$
• D. $4$

Answer 1

Answer: (B)

$2$

Consider the first equation.
Let the roots be $\alpha ,\beta$.
Then the equation becomes
$x^2-(\alpha +\beta )x+\alpha \beta =0$.
Comparing with the given equation
$x^2-6x+a=0$ we get
$\alpha \beta =a$ and $\alpha +\beta =6$ ...(i)
Similarly let the roots of the second equation be $\alpha ,\delta$.
Hence
$x^2-(\alpha +\delta )+\alpha .\delta =0$
Comparing with the given equation $x^2-cx+6=0$
Hence
$\alpha +\delta =c$ and $\alpha .\delta =6$ ...(ii)
Now from i and ii we know that
$\alpha .\beta =a$ and $\alpha .\delta =6$
$\frac{\alpha .\beta }{\alpha .\delta }=\frac{a}{6}$
$\frac{\beta }{\delta }=\frac{a}{6}$.
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
$\frac{\beta }{\delta }=\frac{4}{3}$
Or
$\frac{a}{6}=\frac{4}{3}$
$a=8$
Hence the first equation becomes
$x^2-6x+8=0$
$(x-4)(x-2)=0$
$x=2,4$.
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 $\alpha =2$.
And
$\alpha .\delta =6$
Hence
$\delta =3$.
Therefore the equation becomes
$x^2-(3+2)x+6=0$
$x^2-5x+6=0$.
Hence
$a=8,c=5$ and common root is 2.