The quadratic equations ${\mathrm{x}}^2-6\mathrm{x}+\mathrm{a}=0\text{ and }{\mathrm{x}}^2-\mathrm{cx}+6=0$ have one root $\mathrm{\alpha }$ in common. The other roots of the first and second equations are integers in the ratio 4: 3. Then equation ${\mathrm{x}}^2-3\mathrm{x}-4=0$ has

Let equation ${\mathrm{x}}^2-6\mathrm{x}+\mathrm{a}=0$ have roots $\mathrm{\alpha }\text{ and }\mathrm{\beta }$

Also, let equation ${\mathrm{x}}^2-\mathrm{cx}+6=0$ have roots $\mathrm{\alpha }\text{ and }\mathrm{\gamma }$

Given $\frac{\mathrm{\beta }}{\mathrm{\gamma }}=\frac{4}{3}\Rightarrow \frac{\mathrm{\alpha \beta }}{\mathrm{\alpha \gamma }}=\frac{4}{3}\Rightarrow \frac{\mathrm{a}}{6}=\frac{4}{3}\Rightarrow \mathrm{a}=8$

First equation is ${\mathrm{x}}^2-6\mathrm{x}+8=0$

$\begin{array}{l}\Rightarrow (\mathrm{x}-2)(\mathrm{x}-4)=0\\ \Rightarrow \mathrm{x}=2,4\end{array}$

If $\mathrm{\beta }=4,\text{ then }\mathrm{\gamma }=\frac{3\mathrm{\beta }}{4}=3$

If $\mathrm{\beta }=2\text{, then }\mathrm{\gamma }=\frac{3\mathrm{\beta }}{4}=\frac{3}{2}$, which is not an integer.

Now, ${\mathrm{x}}^2-3\mathrm{x}-4=0$

$\begin{array}{l}\Rightarrow (\mathrm{x}-4)(\mathrm{x}+1)=0\\ \Rightarrow \mathrm{x}=-1,4\end{array}$

Clearly, one root is less than 2 and other is more than2