Question

The number of values of $\alpha \in N$ for which the the quadratic equation $6x^2-11x+\alpha =0$ has rational roots are:

• A. $2$
• B. $1$
• C. $4$
• D. $3$

Answer 1

The correct option is D $3$

Given equation is $6x^2-11x+\alpha =0$
Using discriminant method, we get the roots as:
$x=\frac{-b\pm \sqrt{D}}{2a}$
Now, If roots are rational numbers, then discriminant should be a perfect square.
$\Rightarrow D=b^2-4ac=121-24\alpha$
$D>0\Rightarrow 121-24\alpha >0\Rightarrow \alpha <\frac{121}{24}$

$\because \alpha \in N\Rightarrow \alpha \in \{1,2,3,4,5\}$

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.
$\alpha =3,D=121-24\left(3\right)=49$ which is a perfect square.
$\alpha =4,D=121-24\left(4\right)=25$ which is a perfect square.
$\alpha =5,D=121-24\left(5\right)=1$ which is a perfect square.

Thus, the acceptable values of $\alpha$ are: $3,4,5$
Hence, the number of possible values of $\alpha$ is $3.$