Question

Find the number of all possible positive integral values of α for which the roots of the equation, $6{{x}^{2}}-11x+\alpha =0$ are rational numbers
$\begin{align}
  & \text{a) 2} \\
 & \text{b) 5} \\
 & \text{c) 3} \\
 & \text{d) 4} \\
\end{align}$

Answer 1

Hint: Now we know that the condition for rational roots is that the discriminant is a perfect square. Discriminant of any quadratic equation $a{{x}^{2}}+bx+c=0$ is given by ${{b}^{2}}-4ac$ . hence we can find the discriminant of the given quadratic equation $6{{x}^{2}}-11x+\alpha =0$ and find α using the condition discriminant is a perfect square

Complete step-by-step answer:
Now consider the equation $6{{x}^{2}}-11x+\alpha =0$ . We want to find the value of α such that the roots of the quadratic equation are rational. We know that the condition for rational roots is the discriminant of the quadratic equation is a perfect square.
Hence let us first find the discriminant of the equation $6{{x}^{2}}-11x+\alpha =0$ .
Now we know that discriminant of any quadratic equation $a{{x}^{2}}+bx+c=0$ is given by ${{b}^{2}}-4ac$ .
Hence the discriminant of the equation $6{{x}^{2}}-11x+\alpha =0$ is ${{\left( -11 \right)}^{2}}-4(6)\alpha $, let the discriminant be D.
$D=121-24\alpha $
Now we want D to be a perfect square.
Let us check the values of α such that this is true.
We will not consider α = 0, as only positive values of α is to be considered according to question
Now let us check the value of D for α = 1
D = 121 – 24(1) = 121 – 24 = 97. Now 97 is not a perfect square hence α is not equal to 1.
Now let us check the value of D for α = 2
D = 121 – 24(2) = 121 – 48 = 73 . Now 49 is not a perfect square hence α is not equal to 2
Now let us check the value of D for α = 3
D = 121 – 24(3) = 49. We know 49 is square of 7 hence a perfect square, hence α can be 3
Now let us check the value of D for α = 4
D = 121 – 24(4) = 121 – 96 = 25. We know 25 is also a perfect square hence α can be 4
Now let us check the value of D for α = 5
D = 121 – 24(5) = 1. Now 1 is also a perfect square as $\sqrt{1}=1$. Hence α = 5 is also a possible value
Now we can see that the value of D is negative for all α > 5. For example at α = 6 we get
D = 121 – 24(6) = - 23 Hence those values will give imaginary roots
Hence the only possible values of α for rational roots are α = 3, 4, 5
Hence the number of values of α for which the roots of a given quadratic equation are rational is 3.

So, the correct answer is “Option C”.

Note: Note that in questions it is asked rational numbers and not real numbers. The condition for roots to be real is discriminant should be greater than or equal to 0. But the condition for the roots to be rational is the discriminant must be a perfect square.