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:

To quickly solve quadratic equation root-nature problems, try substituting the options directly into the equation.
With $p = -5$, the equation factors cleanly into $(x+2)(x+3) = 0$, giving integer roots immediately!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a quadratic equation $x^2 - px + 6 = 0$.
We need to determine the value of the parameter $p$ from the given choices that makes the roots of this equation rational.

Step 2: Key Formula or Approach:
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.
From the given equation:
\[ a = 1, \quad b = -p, \quad c = 6 \]
The discriminant is:
\[ D = (-p)^2 - 4(1)(6) = p^2 - 24 \]
We will evaluate this discriminant for each option of $p$ to find which one results in a perfect square.

Step 3: Detailed Explanation:

• Write down the expression for the discriminant:
\[ D = p^2 - 24 \]

• Test the options one by one:
- Option (A): $p = 1$
\[ D = (1)^2 - 24 = 1 - 24 = -23 \]
Since the discriminant is negative, the roots are imaginary.
- Option (B): $p = -5$
\[ D = (-5)^2 - 24 = 25 - 24 = 1 \]
Since $1$ is a perfect square ($1 = 1^2$), the roots of the equation will be rational.
Let us verify the roots of the quadratic equation with $p = -5$:
\[ x^2 - (-5)x + 6 = x^2 + 5x + 6 = 0 \]
\[ (x+2)(x+3) = 0 \implies x = -2 \text{ and } x = -3 \]
Both roots are rational numbers, confirming this choice.
- Option (C): $p = 25$
\[ D = (25)^2 - 24 = 625 - 24 = 601 \]
Since $601$ is not a perfect square, the roots are irrational.
- Option (D): $p = \sqrt{5}$
\[ D = (\sqrt{5})^2 - 24 = 5 - 24 = -19 \]
Since the discriminant is negative, the roots are imaginary.


Step 4: Final Answer:
The value of $p$ for which the roots of the equation are rational is $-5$.