Question:

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

Updated On: July 22, 2025
  • 2
  • 5
  • 3
  • 4
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The Correct Option is C

Solution and Explanation

$ 6x^2 - 11x + \alpha = 0$ given roots are rational $\Rightarrow \; D$ must be perfect square $\Rightarrow \; 121 - 24\alpha = \lambda^2$ $\Rightarrow \; $ maximum value of $\Rightarrow \; $ is 5 $\alpha = 1 \Rightarrow \; \lambda \notin I$ $\alpha = 2 \Rightarrow \; \lambda \notin I$ $\alpha = 3 \Rightarrow \; \lambda \notin I \; \; \; \; \; \Rightarrow 3 $ integral values $\alpha = 4 \Rightarrow \; \lambda \notin I$ $\alpha = 5 \Rightarrow \; \lambda \notin I$
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Concepts Used:

Quadratic Equations

A polynomial that has two roots or is of degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b, and c are the real numbers

Consider the following equation ax²+bx+c=0, where a≠0 and a, b, and c are real coefficients.

The solution of a quadratic equation can be found using the formula, x=((-b±√(b²-4ac))/2a)

Two important points to keep in mind are:

  • A polynomial equation has at least one root.
  • A polynomial equation of degree ‘n’ has ‘n’ roots.

Read More: Nature of Roots of Quadratic Equation

There are basically four methods of solving quadratic equations. They are:

  1. Factoring
  2. Completing the square
  3. Using Quadratic Formula
  4. Taking the square root