Question

The number of all possible positive integral values of $\alpha$ for which the roots of the quadratic equation, $6x^2-11x+\alpha =0$ are rational numbers is :

• A. $3$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is A $3$

Given equation is $6x^2-11x+\alpha =0$
Using discriminant method,
$x=\frac{-b\pm \sqrt{D}}{2a}$
If roots are rational numbers, then discriminant should be a perfect square.
$D=b^2-4ac$
$\Rightarrow$$D=121-24\alpha$
So, If $\alpha =1$, $D=121-24=97$, which is not a perfect square.
$\alpha =2$, $D=121-24\left(2\right)=73$, which is not a perfect square.
$\alpha =3$, $D=121-24\left(3\right)=49$, it is a perfect square.
$\alpha =4$, $D=121-24\left(4\right)=25$, it is a perfect square.
$\alpha =5$, $D=121-24\left(5\right)=1$, it is a perfect square.
If we take greater values of $\alpha$, then $D$ will become negative.

So, number of possible values of $\alpha$ is $3$.