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:

For multiple-choice questions, directly substitute the options into the equation to see which one leads to simple factorization.
Substituting $p = -5$ gives $x^2 + 5x + 6 = 0$, which instantly factors into $(x+2)(x+3) = 0$!

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 find the value of $p$ from the given options such that the roots of this equation are rational numbers.

Step 2: Key Formula or Approach:
The roots of a quadratic equation $ax^2 + bx + c = 0$ with rational coefficients are rational if and only if the discriminant $D = b^2 - 4ac$ is a perfect square of a rational number.
Here, $a = 1$, $b = -p$, and $c = 6$.
The discriminant is:
\[ D = (-p)^2 - 4(1)(6) = p^2 - 24 \]
We will check each option to see which one makes $D$ a perfect square.

Step 3: Detailed Explanation:

• Evaluate the discriminant for each option of $p$:
- Option (A): $p = 1$
\[ D = 1^2 - 24 = -23 \]
Since $D \lt 0$, the roots are imaginary (complex) and not rational.
- Option (B): $p = -5$
\[ D = (-5)^2 - 24 = 25 - 24 = 1 \]
Since $D = 1$ is a perfect square ($1 = 1^2$), the roots will be rational.
Let us verify by solving the equation:
\[ x^2 - (-5)x + 6 = x^2 + 5x + 6 = 0 \implies (x+2)(x+3) = 0 \implies x = -2, -3 \]
Since $-2$ and $-3$ are rational, this choice is correct.
- 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 $D \lt 0$, the roots are imaginary.


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