Question

The value of \(p\) for which roots of the quadratic equation \(x^{2} - px + 6 = 0\) are rational, is

• A. \(1\)
• B. \(-5\)
• C. \(25\)
• D. \(\sqrt{5}\)

Hint:

Roots are rational only if \(D \ge 0\) and \(D\) is a perfect square. If \(D\) is not a perfect square (like \(2\), \(3\), etc.), roots are irrational. If \(D<0\), roots are non-real.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a quadratic equation \(ax^2 + bx + c = 0\) with rational coefficients, the roots are rational if and only if the discriminant \(D = b^2 - 4ac\) is a perfect square of a rational number.
Step 2: Key Formula or Approach:
Discriminant formula: \[ D = b^2 - 4ac \]
Given equation: \(x^2 - px + 6 = 0\)
Here, \(a = 1, b = -p, c = 6\).
Step 3: Detailed Explanation:
Substitute the values into the discriminant formula:
\[ D = (-p)^2 - 4(1)(6) \]
\[ D = p^2 - 24 \]
We test the given options for \(p\) to see which makes \(D\) a perfect square:
(A) If \(p = 1\): \(D = 1^2 - 24 = -23\) (Not a perfect square; roots are non-real).
(B) If \(p = -5\): \(D = (-5)^2 - 24 = 25 - 24 = 1\). Since \(1\) is a perfect square (\(1^2\)), the roots are rational.
(C) If \(p = 25\): \(D = 25^2 - 24 = 625 - 24 = 601\) (Not a perfect square).
(D) If \(p = \sqrt{5}\): \(D = (\sqrt{5})^2 - 24 = 5 - 24 = -19\) (Not a perfect square; roots are non-real).
Step 4: Final Answer:
The value of \(p\) is \(-5\).