The quadratic equations $x^2-6 x+a=0$ and $x^2-c x+6=0$ have one root in common. The other roots of the…
The quadratic equations $x^2-6 x+a=0$ and $x^2-c x+6=0$ 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
1
4
3
2
Solution
Let $\alpha$ and $4 \beta$ be roots of $x^2-6 x+a=0$ and $\alpha, 3 \beta$ be the roots of $x^2-c x+6=0$, then $\alpha+4 \beta=6$ and $4 \alpha \beta=a$
$\alpha+3 \beta=c$ and $3 \alpha \beta=6$.
We get $\alpha \beta=2 \Rightarrow a=8$
So the first equation is $x^2-6 x+8=0 \Rightarrow x=2,4$
If $\alpha=2$ and $4 \beta=4$ then $3 \beta=3$
If $\alpha=4$ and $4 \beta=2$, then $3 \beta=3 / 2 \quad$ (non-integer)
$\therefore$ common root is $x=2$.